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MathematicalEconomics

Justin Raymond S. EloriagaDe La Salle UniversitySchool of Economics

Term 1Academic Year 2020 - 2021

For my Dad up thereMay the Force be with you, always

Preface

"In this world, everything is governed by balance. There’s what you stand to gain and what you stand to lose.And when you think you’ve got nothing to lose, you become overconfident" - The Professor

La Casa de Papel/Money Heist

In mid 2019 right before graduation after I had secured my job search, I had bummed about just doing my hobbiesand binged watch countless Netflix shows to make up for lost ground while grinding my graduate thesis. Since Iwould end up working for the central bank, I was recommended to watch a show called Money Heist by a friend(who was of course not from the central bank). The premise of the show was your typical drama/crime/thrillerwith a lot of well written characters, the names of which remind me of my countless travels and misadventuresright before 2020 and the pandemic. I, of course, related to the professor character (being one and all) andwas intrigued by the line he said above. As in anything in economics, balance is often desired. We desire thisconstruct of equilibrium. But he posits something very game theory-ish in that when the odds seem in ourfavor, we can become overconfident. I guess this best describes my affinity with mathematics in the context ofeconomics. On the one hand having gone through many courses involving math, you’d think you’d be good atit. And yet, till today, I still find myself dumbfounded by simple problems. But trust me when I say this, it isdefinitely okay. Do not fret too much, it is a learning process to learn math. Many may deem it a daily struggle.However, I will want to prove to you that it is an indispensable tool in the economist’s tool kit, one that is oftencharacterized as what differentiates us from other social sciences.

This is a compilation of my notes in Mathematical Economics. I have constructed it in such a waythat it approaches the subject matter rigorously but in a more paletable and approachable manner. Asthis is just my personal work, there are bound to be errors to which I apologize in advance. Videos onvarious lessons can be found on my YouTube channel, just search Justin Eloriaga. Feel free to like, share,and subscribe as the channel also includes other courses which you might eventually end up taking. Thiswork was primarily based on the notes of my collegues, namely, Ailyn Ang Shi and Evan Li Liao which Isupplemented with a deeper discussion in some areas and more examples. The main references used areChaing and Wainwright, Sydsaeter and Hammond, and Schaum

Justin S. Eloriaga Mathematical Economics

Contents

1 Differentiation 61.1 Rates of Change, Derivative, and Tangent Line . . . . . . . . . . . . . . . . . . . . . . . 6

1.1.1 Rates of Change . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.1.2 Tangent Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.1.3 Differentiability and Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.2 Rules of Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.2.1 Constant Function Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.2.2 Linear Function Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.2.3 Power Function Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.2.4 Constant Multiple Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.2.5 Sum and Difference Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.2.6 Product Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.2.7 Quotient Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.2.8 Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.2.9 Derivatives of Exponential and Logarithmic Functions . . . . . . . . . . . . . . . 111.2.10 Implicit Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.3 Higher Order Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131.3.1 Second Order Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131.3.2 Continously Differentiable Function . . . . . . . . . . . . . . . . . . . . . . . . . . 141.3.3 Implication of the Second Derivative . . . . . . . . . . . . . . . . . . . . . . . . . 14

1.4 Differential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151.4.1 Economic Application using National Income Accounting . . . . . . . . . . . . . 151.4.2 Property of a Differential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

1.5 Differentiation with Several Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171.5.1 Partial Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171.5.2 Higher-Order Partial Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . 171.5.3 Marginal Functions in Economics . . . . . . . . . . . . . . . . . . . . . . . . . . . 181.5.4 Total Differentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

2 Applications of Differentiation 202.1 Taylor Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

2.1.1 Linear Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202.1.2 Quadratic Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

2.2 Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212.2.1 Local Maxima and Minima . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212.2.2 The First-Derivative Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

2.3 Concavity and Convexity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242.3.1 Strict Concavity and Convexity . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242.3.2 Weak Concavity and Convexity . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242.3.3 Slope and Shape Using the First and Second order Derivatives . . . . . . . . . . 25

2.4 The Second Derivative Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252.4.1 Inflection Point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

2.5 Global Optimum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282.5.1 Weierstrauss’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282.5.2 Rolle’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

2.6 Profit Maximization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

3 Integration 313.1 Anti-differentiation and Indefinite Integrals . . . . . . . . . . . . . . . . . . . . . . . . . 31

3.1.1 The Antiderivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 313.1.2 Power Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 313.1.3 Constant Number Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

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3.1.4 Factorization and Separation Rules . . . . . . . . . . . . . . . . . . . . . . . . . . 313.1.5 Logarithmic and Exponential Rules . . . . . . . . . . . . . . . . . . . . . . . . . . 32

3.2 Area Under a Curve and Definite Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . 333.2.1 Reimann Sum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 343.2.2 Definite Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 343.2.3 Fundamental Theorems of Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . 343.2.4 Additional Rules on Definite Integrals . . . . . . . . . . . . . . . . . . . . . . . . 36

3.3 Integration by Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 373.4 Integration by Parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

3.4.1 Integration by Parts of Indefinite Integrals . . . . . . . . . . . . . . . . . . . . . . 383.4.2 Integration by Parts for Definite Integrals . . . . . . . . . . . . . . . . . . . . . . 38

3.5 Integration by Using Partial Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 393.6 Integration by Using Partial Fractions with Repeated Factors . . . . . . . . . . . . . . . 403.7 Improper Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 413.8 Economic Applications of Integrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

3.8.1 Finding Total Cost Functions from Marginal Cost Functions . . . . . . . . . . . . 433.8.2 Finding Total Revenue Functions from Marginal Revenue Functions . . . . . . . 433.8.3 National Income, Consumption, and Savings . . . . . . . . . . . . . . . . . . . . . 443.8.4 Investment, Capital Formation, and Capital Accumulation . . . . . . . . . . . . . 443.8.5 Consumer, Producer, and Social Surpluses . . . . . . . . . . . . . . . . . . . . . . 46

4 Matrix Algebra 484.1 Introduction to Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

4.1.1 Special Forms of Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 484.2 Matrix Operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

4.2.1 Matrix Addition and Subtraction . . . . . . . . . . . . . . . . . . . . . . . . . . . 504.2.2 Scalar Multiplication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 504.2.3 Matrix Multiplication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 504.2.4 Properties of Matrix Operations . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

4.3 Solving Systems of Linear Equations using Matrices . . . . . . . . . . . . . . . . . . . . 524.3.1 Matrix Algebra and the Matrix Form . . . . . . . . . . . . . . . . . . . . . . . . . 524.3.2 Defining Determinants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 534.3.3 Solving for Determinants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 544.3.4 Basic Properties of Determinants . . . . . . . . . . . . . . . . . . . . . . . . . . . 564.3.5 Defining the Inverse of a Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . 574.3.6 Solving for the Inverse of a Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . 574.3.7 Solving a System of Equations by Matrix Inversion . . . . . . . . . . . . . . . . . 594.3.8 Solving a System of Equations using Cramer’s Rule . . . . . . . . . . . . . . . . . 60

5 Economic Applications of Matrix Algebra and Comparative Statics 615.1 Single-Commodity Competitive Market . . . . . . . . . . . . . . . . . . . . . . . . . . . . 615.2 National Income Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 625.3 Closed Economy IS-LM Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 625.4 Leontief Input-Output Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

5.4.1 The Leontief Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 645.4.2 Demand for Primary Inputs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

5.5 Comparative Statics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 675.5.1 The Jacobian Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

6 Multivariate Unconstrained Optimization 706.1 Topological Underpinnings of the First Derivative . . . . . . . . . . . . . . . . . . . . . . 706.2 Second Derivative Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

6.2.1 Hessian Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 726.2.2 Definiteness and Principal Minors . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

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6.2.3 Determining the Multivariate SOC . . . . . . . . . . . . . . . . . . . . . . . . . . 746.2.4 Concavity and Convexity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

6.3 Comparative Statics and Specific Functional Forms . . . . . . . . . . . . . . . . . . . . . 75

7 Constrained Optimization 767.1 Optimization with One Equality Constraint: The Two-Variable Case . . . . . . . . . . . 76

7.1.1 The Substitution Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 777.1.2 The Lagrange Multiplier Method . . . . . . . . . . . . . . . . . . . . . . . . . . . 777.1.3 The Bordered Hessian Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

7.2 Optimization with One Equality Constraint: The n variable case . . . . . . . . . . . . . 80

8 Course Problem Sets 828.1 Problem Set 1: Single Variable Differentiation . . . . . . . . . . . . . . . . . . . . . . . . 838.2 Problem Set 2: Multivariate Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . 858.3 Problem Set 3: Economic Applications of Differentiation . . . . . . . . . . . . . . . . . . 868.4 Problem Set 4: Techniques of Integration . . . . . . . . . . . . . . . . . . . . . . . . . . 878.5 Problem Set 5: Deep Dive on Matrix Algebra . . . . . . . . . . . . . . . . . . . . . . . . 898.6 Problem Set 6: Economic Applications of Matrix Algebra . . . . . . . . . . . . . . . . . 928.7 Problem Set 7: Unconstrained Optimization . . . . . . . . . . . . . . . . . . . . . . . . . 948.8 Problem Set 8: Constrained Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . 96

9 Course Syllabus 989.1 Course Description . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 989.2 Main References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99

9.2.1 Required References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 999.2.2 Other References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99

9.3 Grading System and Requirements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 999.4 Lectures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 999.5 Course Plan . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1009.6 Course Guidelines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1019.7 Contact and Consultation Hours . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101

5


1 Differentiation

In this section we will discuss the rudiments of differentiation in economic. We will tackle the underlyingbasis first which means going through the concept of a derivative and how it is articulated mathemat-ically. We then move on to the various techniques for differentiation and then move to multivariatedifferentiation.

1.1 Rates of Change, Derivative, and Tangent Line

1.1.1 Rates of Change

The average rate of change of f over the interval [a, a+ h], or the difference quotient is

∆y

∆x=f(a+ h)− f(a)

h(1)

In this equation, ∆x = h is called the increment of x, and ∆y = f(a+ h)− f(a) is the respectiverate of change in y. We can modify this slightly by taking the instantaneous rate of change of f at aand is represented by the limit

limh→0

f(x+ ∆x)− f(x)

∆x(2)

The derivative of a function f is a function, denoted as f ′, such that its value a a number x inthe domain of f is given by the form if the limit exists.

Theorem 1.1. Provided that the limit exists, the derivative is given as f ′(x) = lim∆x→0f(a+h)−f(a)

h

To understand this concept, let us picture this scenario. Suppose a car, initially at rest, moved alonga straight road in the same direction and its motion was monitored during the first 60 seconds. Thefollowing data were recorded, namely, t which is the time in seconds from the start, and s which is thedistance travelled in feet from the start.

t (seconds) 15 20 25 30 35 40 45

s (feet) 168 300 468 675 918 1200 1518

Consider the time interval 15 ≤ t ≤ 30. The average speed during this interval is

∆s

∆t=

675− 168

30− 15= 33.800

feetsecond

This figure which is the average speed gives us a rough idea of how fast the car is running duringan interval of time. But it does not give us the speed at an instant in time. To illustrate, see the tablebelow.

Interval Length Average Speed Beforet = 30

Average Speed Aftert = 30

15 seconds (675− 168)/15 = 33.8 (1518− 675)/15 = 56.210 seconds (675− 300)/10 = 37.5 (1200− 675)/10 = 52.55 seconds (675− 468)/5 = 41.4 (918− 675)/5 = 48.61 second (675− 630)/1 = 45.0 (720− 675)/1 = 45.0

The average speeds before and after the 30th second tend toward each other as the size of the intervalbecomes smaller and smaller. The common value approached by these average speeds as the intervalshrinks to zero is the instantaneous speed at t = 30.

6


We say that f is a differentiable function and call f ′(x) as the derivative of f with respect tox. The derivative allows us to compute for the rate of change at any number within the domain of thefunction. We refer to equation one as the difference quotient.

Say we want to determine the derivative of the function f(x) = 4x2 + 2. To do this, we can merelyuse the difference quotient.

f ′(x) = lim∆x→0

f(x+ ∆x)− f(x)

∆x= lim

∆x→0

4(x+ ∆x)2 + 2− (4x2 + 2)

∆x

We can simplify this as

f ′(x) lim∆x→0

4(x2 + 2x∆x+ ∆x2) + 2− (4x2 + 2)

∆x


4x2 + 8x∆x+ 4∆x2 + 2− 4x2 − 2

∆x


8x∆x+ 4∆x2

∆x= lim

∆x→08x+ 4∆x

We evaluate this limit and we find that

f ′(x) = 8x

Find the derivatives of the following functions using the difference quotient.

1. f(x) = 3x2 + 1

2. f(x) = 1x , x 6= 0

Note that our notation for differentiation may take four different forms. These are the LeibnizNotation dy

dx , the Euler notation Dxf , the Newton Notation y, or the Lagrange Notation f ′(x).

1.1.2 Tangent Line

The tangent line to the curve y = f(x) at the point P = (a, f(a)) is the line through P with the slopeprovided that the limit below exists.

m = limh→0

f(a+ h)− f(a)

(a+ h)− a= lim

h→0

f(a+ h)− f(a)

h(3)

Hence, the derivative is simply the slope of the tangent to the curve which, by definition, is theslope of the curve.

Figure 1: Slope of the Tangent Line

7


If we write x = a+ h, then h = x− a, and h approaches 0 if and only if x approaches a. Therefore,an equivalent way of stating the slope of the tangent line is:

f ′(a) = limx→a

f(x)− f(a)

x− aThe slope of any one curvilinear function is not constant. This may differ at different points on the

curve as in figure 1. A tangent line is a straight line that touches a curve at only one point. Measuringthe slope of a function may require different tangent lines.

1.1.3 Differentiability and Continuity

For a function to be differentiable at some x∗, the tangent to the curve at x∗ has to be the sametangent whether x∗ is approaches from the left or from the right. That is, there should be no abruptchange in slope in the neighborhood of x∗, or in other words, the curve should be smooth around x∗.

Theorem 1.2. If f is differentiable at a, then f is continous at a

The theorem implies that differentiability implies continuity or that continuity is a necessary con-dition for differentiability. Note that the converse is not generally true as there are functions that arecontinuous but are not differentiable at certain points.

(a) f(x) = |x| (b) f(x) = x13

Figure 2: Functions Illustrating Theorem 1.2

1.2 Rules of Differentiation

We formally refer to differentiation as the process of finding the derivative of a function. This involvesnothing more complicated than applying a few basic rules.

1.2.1 Constant Function Rule

If c is a constant, and if f(x) = c for all x, then the derivative of that function is zero.

f ′(x) = 0

1.2.2 Linear Function Rule

If we have a linear function f(x) = mx + b, the derivative of the linear function is merely m. Thisstems from the derivative of a variable raised to the first power is always equal to the coefficient of thevariable, while the derivative of a constant is simply zero.

Say we were given with the linear function f(x) = 7x+ 49

f ′(x) = 7

8


1.2.3 Power Function Rule

If n is a positive integer, and if we have the function f(x) = xn, then it follows that the derivative isgiven as

f ′(x) = nxn−1

For example, say we are given with f(x) = 2x2. The derivative of this is simply

f ′(x) = 2 · 2x2−1 = 4x

1.2.4 Constant Multiple Rule

If f is a differentiable function, c is a constant and g is a function defined by g(x) = cf(x), then itfollows that

g′(x) = cf ′(x)

For example, take g(x) = 8x2 where c is obviously 8 and f(x) = x2. We know from the Powerfunction rule that the derivative of f(x) is f ′(x) = 2x. Hence, getting the derivative of g(x) is straight-forward

g′(x) = 8 · f ′(x) = 8 · 2x = 16x

1.2.5 Sum and Difference Rules

If f and g are both differentiable functions, and if h is a function defined by h(x) = f(x)± g(x), thenthe derivative of h′(x) is merely the sum or difference of the derivatives of f(x) which is f ′(x) and g(x)which is g′(x).

h′(x) = f ′(x)± g′(x)

For example., say we had a function h(x) = x2 + 2x5 + 34x

4. The derivative of this function h′(x) ismerely the derivative of each term with respect to x.

h′(x) =dh(x)

dx(x2) +

dh(x)

dx(2x5) +

dh(x)

dx

(3

4x4

)= 2x+ 10x4 + 3x3

Using the rules you know so far, take the derivative of the following functions

1. f(x) = x8 + 12x5 − 4x4 + 10x3 − 6x+ 5

2. f(x) = 5x3

3. f(x) = 1x4− x4

4. f(x) = x0.5 − 2x− 1

5. The area A of a circular disc of radius r is given by A = πr2. Calculate the derivative of Awith respect to r and give its geometric interpretation.

9


Marginal Cost

In economics, we can of course measure or quantify cost, and, in general, cost is expressed as somefunction of quantity. Given a cost function C(q) we can get the derivative of the cost with respect toquantity produced to obtain the marginal cost function, which we denote as C ′(q) which gives usthe rate of change in cost per unit produced.

For example, suppose that the total cost in pesos incurred each week by a manufacturing firm for qunits of a product is given by the total cost function

C(q) = 8000 + 200q − 0.2q2

The marginal cost can then be obtained by taking the derivative with respect to q that will yield

C ′(q) =dC(q)

dq= 200− 0.4q

Therefore, each additional q produced will yield an increase in the total cost equal to 200 − 0.4q,ceteris paribus.

1.2.6 Product Rule

If f and g are both differentiable functions, and if h is a function defined by h(x) = f(x)g(x), then thederivative of h(x) which is h′(x) can be obtained using the form below

h′(x) = f ′(x)g(x) + g′(x)f(x) (4)

For example, say you are given with the function h(x) = (x+ 5)(x2 + 1). From here, we can assignf(x) = (x+ 5) and g(x) = (x2 + 1).To derive this function, simply do

h′(x) = (1)︸︷︷︸f ′(x)

(2x+ 1)︸︷︷︸g(x)

+ 2x︸︷︷︸g′(x)

(x+ 5)︸︷︷︸f(x)

h′(x) = (2x+ 1) + 2x(x+ 5) = 2x+ 1 + 2x2 + 10x = 2x2 + 10x+ 1

Marginal Revenue Function

The revenue function is defined as R(qj) = p(Q)qj for a perfectly competitive market. If the firm is amonopolist, we have Q = qj = q. We define the marginal revenue as the revenue earned by a firmfor producing an additional unity of a good. The marginal revenue function is the rate of change inrevenue per unit of the good sold, and is obtained by differentiating the revenue function with respectto q. To obtain this, we can merely use the product rule (for cases of perfect competition)

MRPC = R′(q) = p′(q)q + p(q)

1.2.7 Quotient Rule

If f and g are both differentiable functions, and if h is a function defined by h(x) = f(x)g(x) , then it holds

that

h′(x) =g(x)f ′(x)− f(x)g′(x)

(g(x))2(5)

For example, say we were given with the function h(x) = x2x−4 . First, lets treat f(x) = x and

g(x) = 2x− 4. Following the quotient rule

h′(x) =(2x− 4)(1)− (x)(2)

(2x− 4)2=

2x− 4− 2x

4x2 − 16x+ 16=

−4

4x2 − 16 + 16

10


Get the derivative of f(x) = x2

(x2+3)with respect to x

1.2.8 Chain Rule

If f and g are both differentiable functions, and if h is a function defined by h(x) = f(g(x)), then

h′(x) = f ′(g(x))g′(x) (6)

For example. say you were asked to get the derivative of h(x) = (3x + 1)2 with respect to x usingthe Chain Rule. Say that we think of 3x+ 1 as g(x). To solve using the chain rule, this translates into

h′(x) = 2 · (3x+ 1)2−1 · dg(x)

dx(3x+ 1)

h′(x) = 2(3x+ 1) · (3) = 6(3x+ 1) = 18x+ 6

Alternatively, we can factor this binomial out and take the derivative term by term instead of usingthe chain rule since the power is not to high (not so complicated to get). When we factor, we get thath(x) = 9x2 + 6x+ 1. Taking the derivative of this function yield

dh(x)

dx= h′(x) = 18x+ 6

Find the derivative of the following functions with respect to x using the chain rule.

1. f(x) = (x3 + 2)5

2. f(x) =(

1x + 1

)2

1.2.9 Derivatives of Exponential and Logarithmic Functions

If f(x) = ex, then the derivative of f(x) is merely the original function.

f ′(x) = ex (7)

For example, say we were given with the function f(x) = e2x−3. To solve this, we need to apply therule above and also chain rule.

f ′(x) = e2x−3 · ddx

(2x− 3) = e2x−3 · 2 = 2e2x−3

If f(x) = lnx, then the derivative of the function is merely 1x

f ′(x) =1

x(8)

Consider the function f(x) = ln(x2 + 3x+ 7), deriving this needs the rule above and also the chainrule.

f ′(x) =1

x2 + 3x+ 7· ddx

(x2 + 3x+ 7) =2x+ 3

x2 + 3x+ 7

If a is a constant number and f(x) = ax, then

f ′(x) = ax ln a (9)

Say we were given with y = 7x, we can derive this function with respect to x and yields

y′ = 7x ln(7)

11


Find the derivative of the following functions

1. y = e−x

2. y = 3e1−x

3. y = xe1−x

4. y = ex2

5. y = ln(ex + e−x)

6. y = 12x+ex

1.2.10 Implicit Differentiation

When variables y and x are related by an equation of the form F (x, y) = 0, we say that y is an implicitfunction of x. With implicit functions, it may be difficult to express y as an explicit function of x, ory = f(x). Thus, implicit differentiation may be useful.

For instance, say we had the function y3 + y2 + x2 = 2

Differentiating term by term with respect to x, we get

3y2 dy

dx+ 2y

dy

dx+ 2x = 0

Rearranging

dy

dx(3y2 + 2y) = −2x

dy

dx=

−2x

3y2 + 2y

Find dydx by implicit differentiation

1. x2 + y2 − 3 = 0

2. x2 + y2 = xy

3. y + x2y − 1 = 0

Growth Rate of a Variable

We define the growth rate of a variable x(t) that varies across one time period as

gx =x(t+ 1)− x(t)

x(t)(10)

In general, we compute the growth rate of x(t) across several time periods as

gx =x(t+ ∆t)− x(t)

∆t· 1

x(t)

Letting h approach zero, or computing the growth rate of x(t) over a very small time period, weredefine the growth rate of a variable as

lim∆t→0

x(t+ ∆t)− x(t)

∆t· 1

x(t)=x′(t)

x(t)≈ gx

12


Elasticity

From basic microeconomics, the elasticity of demand with respect to price is defined as

ε = −q2−q1q1

p2−p1p1

= −∆qq

∆pp

In the form above, ∆q is the change in quantity demanded and ∆p is the change in prices. We saythat the good is price inelastic if |ε| < 1, price elastic if |ε| > 1, and unitary inelastic if |ε| = 1/

Rewriting the formula for elasticity yields

ε = −∆qq

∆pp

= −∆q

∆p· pq

Letting the change in prices approach 0, we obtain

−pq

lim∆p→0

∆q

∆p=p

qq′ ≈ ε

However, most microeconomic textbooks will use the format

ε = −pq· dqdp

(11)

For example, suppose that the demand for a commodity is given by q = 120 − 4p where q is thequantity demanded and p is the price. We first determine the price elasticity of demand which is thepoint elasticity of q with respect to p

ε = −pq· dqdp

= −pq· −4

If you were asked to determine and interpret the price elasticity when the price is say p = 10, youwould need to plug this in to the demand function first to get q.

q = 120− 4(10) = 120− 40 = 80

Plugging this in to our elasticity form

ε = −pq· dqdp

= −10

80· −4 =

40

80=

1

2

Since ε = 1/2 < 1, we can sat that the demand for this commodity is price inelastic.

Using the same demand function used, determine and interpret the price elasticity of demand atp = 20 and at p = 30. What do you observe?

1.3 Higher Order Derivatives

1.3.1 Second Order Derivative

If f is a differentiable function, then its derivative f ′ is also a function, so f ′ may have a derivative ofits own, denoted by (f ′)′ = f ′′. This new function is called the second order derivative of f becauseit is the derivative of the derivative of f . We say that f is twice differentiable.

Consider the function y = x3 − 3x2 + 1. We take the first order derivative of this function as

dy

dx= y′ = 3x2 − 6x

13


To get the second order derivative, we take the derivative of this (first-order) derivative as what youcan see below.

d2y

dx2= y′′ = 6x− 6

Get the first and second order derivative of the following functions

1. y = 7x2 + 2x3 + 78

2. y = (2x2 + 2)2

3. y = 2x− 1

1.3.2 Continously Differentiable Function

In more general notation, we have f (n)(x) called the nth order derivative of f with respect to x. If fis a differentiable function and its derivative is a continuous function, we say that f is a continuouslydifferentiable function.

If f s twice differentiable and its second order derivative is also a continuous function, we say thatf is a twice continuously differentiable function.

Derive the following functions as far as you can until you get a derivative equal to zero

1. y = 7x2 + 2x3 + 78

2. y = 9x8 + 8x7 + 7x6 + 6x5 + 5x4 + 4x3 + 3x2 + 2x+ 1

1.3.3 Implication of the Second Derivative

Consider the graph of a function f . From point A to B, the slope is positive but decreasing until itbecomes zero at B. Beyond B, the slope is negative and decreasing up to C. We then say that the graphis concave from A to C. From C to D, the slope is negative but increasing until it becomes zero at D.Beyond D, the slope is positive and increasing up to E. We then say that the graph is convex from Cto E.

(a) Showing Concave and Convexin f (b) Convex Graphs (c) Concave Graphs

Figure 3: Shape of Functions

14


For example, consider the function f(x) = x2 + 2x+ 1. We can obtain the first and second order ofthis function as

f ′(x) = 2x+ 2

f ′′(x) = 2 > 0

To identify on what range the function is concave or convex, we can plug in various values of x,When x > −1, then f ′(x) > 0 which means that the graph on these points is convex coming froma higher slope going to a lower slope. When x = −1, f(x) = 0 which means that the graph is at astationary point in this range. When x < −1, then f ′(x) < 0 which means that the graph at this pointis convex still.

Figure 4: Graph of f(x) = x2 + 2x+ 1

1.4 Differential

If y = f(x), where f is a differentiable function, then the differential dy is defined by the equation

dy = f ′(x)dx

In the form above, f ′(x) is the derivative of y with respect to x and dx = ∆x is the change in x.The process of obtaining the differential of y is called total differentiation. The differential measuresthe estimated change in y given a change in x, that is ∆y ≈ dy.

1.4.1 Economic Application using National Income Accounting

Given a three-sector economy Y = C + I +G where household consumption is determined by a linearconsumption function that depends on income, C(Y ) = α + βY . Further assume that consumptionand investment are exogenous to the model and only government expenditure is endogenous. Theequilibrium output of this basic economy is just merely

Y ∗ = Y ∗(G) =α+ I +G

1− βIf government expenditure changes, we can compute the multiplier effect by totally differentiating

Y ∗.

dY ∗ = Y ∗(G)dG =1

1− βdG

In the expression above, we refer to 11−β as the multiplier effect. To illustrate, suppose that the

consumption function C(Y ) = 5 + 0.5Y and investments are fixed at 50. Then the multiplier is equalto 2. If If initial government expenditures was 100 and an expansionary fiscal policy increases it to 101,then the change in government expenditures is dG = 1. This means that the change in equilibrium

15


output is dY ∗ = 2. Note that the change dY is just the multiplier (which is 2) multiplied to the changein G which was dG. We can test this out by actually computing for this change.

Y ∗(101)− Y ∗(100) =5 + 50 + 101

0.5− 5 + 50 + 100

0.5=

156

0.5− 155

0.5= 312− 310 = 2

1.4.2 Property of a Differential

If y = f(x), where f is a differentiable function, and the change in x is very small, then the differentialof y is close to the actual change in y. That is:

limdx→0

∆y = dy (12)

Graphically, we illustrate the differential as

Figure 5: Illustrating the Differential

In our linear example in the three sector economy, we determined that dy = ∆Y . However, whenfunctions are non-linear, this may not always be the case. We may only get an approximation as indy ≈ ∆Y

Theorem 1.3. As the differential dx gets smaller, the difference between the actual change ∆y and thedifferential dy goes closer and closer to zero.

Proof. Let y = f(x) = x2 and let x = 2

If dx = 0.1 = ∆x, then

∆y = f(2.1)− f(2) = (2.1)2 − (2)2 = 0.41

dy = 2xdx = 2(2)(0.1) = 0.4

∴ ∆y − dy = 0.41− 0.40 = 0.01

If dx = 0.01 = ∆x, then

∆y = f(2.01)− f(2) = (2.01)2 − (2)2 = 0.0401

dy = 2xdx = 2(2)(0.01) = 0.04

∴ ∆y − dy = 0.0001

Thus, as the differential dx becomes smaller, the differential dy gets closer to the actual change∆y

16


1.5 Differentiation with Several Variables

1.5.1 Partial Derivatives

It is myopic to believe that all functions are just functions with one variable. Let z = f(x, y). Thepartial derivative of f with respect to x at (x, y) is a function fx defined by the form below if thelimit exists.

fx(x, y) = lim∆x→0

f(x+ ∆x, y)− f(x, y)

∆x|y=y

Similarly, the partial derivative of f with respect to y at (x, y) is a function fy is defined bythe form below if the limit exists

fy(x, y) = lim∆y→0

f(x+ ∆x, y)− f(x, y)

∆y|x=x

Likewise, we also use the following notations for the partial derivative

fx(x, y) =∂z

∂x=∂f

∂x=

∂

∂xf(x, y)

fy(x, y) =∂z

∂y=∂f

∂y=

∂

∂yf(x, y)

To find the partial derivative with respect to x, that is, fx(x, y), regard y as a constant and differ-entiate f(x, y) with respect to x. To find fy(x, y) or the partial derivative with respect to y, regard xas a constant and differentiate.

For example, say we have f(x, y) = 2x2 + y3 + xy. Differentiate the function with respect to bothx and y. Doing this should be straightforward

fx(x, y) =∂f

∂x= 4x+ y

fy(x, y) =∂f

∂y= 3y2 + x

Obtain the partial derivative with respect to each variable

1. f(x, y) = 3y2ex + y2

2. f(x, y, z) = xyz2 + 3xyz + z3

3. f(α, β) = αεβ1−ε

1.5.2 Higher-Order Partial Derivatives

Since the partial derivative is a function, we may also define its partial derivatives. These are calledsecond-order partial derivatives and are denoted as follows

fxx = (fx)x =∂

∂x·(∂z

∂x

)=∂2z

∂x2

fyy = (fy)y =∂

∂y·(∂z

∂y

)=∂2z

∂y2

fyx = (fy)x =∂

∂x·(∂z

∂y

)=

∂2z

∂x∂y

fxy = (fx)y =∂

∂y·(∂z

∂x

)=

∂2z

∂y∂x

17


We refer to fxx and fyy as the direct second order partial derivatives. We refer to fxy and fyxas the cross partial derivatives.

Theorem 1.4. The cross partial derivatives are necessarily equal to each other (Young’s Theorem)

Instead of showing a formal proof, let us use an example. Consider the function z = f(x, y) =x3 + x2y2 + y3. We obtain the first-order partial derivatives

fx(x, y) = 3x2 + 2xy2

fy(x, y) = 2x2y + 3y2

We obtain the cross partial derivativesfxy = 4xy

fyx = 4xy

Hence, Young’s Theorem holds since the cross partial derivatives are necessarily equal.

Find the second order direct and cross partial derivatives of the following functions. Demonstratethat Young’s Theorem holds.

1. z = f(x, y) = xey

2. z = f(x, y) = 3xy + x5 + y3

1.5.3 Marginal Functions in Economics

Consider a production function with capital and labor as its factors of production, F (K,L). Then thepartial derivative FK(K,L) is called the marginal productivity of capital, which measures the rateof change of production with respect to the amount of capital used with the level of labor held fixed,and FL(K,L) is called the marginal productivity of labor, which measures the rate of change ofproduction with respect to the amount of labor used with the level of capital held fixed.

For example, consider a Cobb-Douglas Production function F (K,L) = AKαLβ . We obtain themarginal product of capital and the marginal product of labor by just getting the first order partialderivative with respect to each input.

∂F (K,L)

∂K= αAKα−1Lβ = fK(K,L) = MPK

∂F (K,L)

∂L= βAKαLβ−1 = fL(K,L) = MPL

Notice that under a reasonable domain where K > 0 and L > 0, then both MPK and MPL aregreater than zero or strictly positive. This is because for the most part, an increase in labor (or capital),holding the other input fixed (ala ceteris paribus), will increase the output F (K,L).

Despite this being positive, there still exists diminishing returns. The law of diminishing returnsstates that after a certain point, when additional units of a variable input are added to fixed inputs,the marginal product of the variable input declines. Given a production function F (K,L) with marginalproductivity of capital and labor, fK and fL, the law of diminishing marginal productivity is exhibitedby mathematically as fKK and fLL are less than zero.

Theorem 1.5. The Law of Diminishing Marginal Productivity requires that for a typical productionfunction, fxx, fjj < 0 ∀x, j > 0

18


From our example, we know that MPK = αAKα−1Lβ = fK(K,L) and MPL = βAKαLβ−1 =fL(K,L). Further note that 0 ≤ α ≤ 1 and 0 ≤ β ≤ 1. Obtaining the second order direct partialderivatives

fKK = (MPK)K = α(α− 1)Kα−2Lβ = (α2 − α)Kα−2Lβ < 0 ∀K,L > 0

fLL = (MPL)L = β(α− 1)KαLβ−2 = (β2 − β)KαLβ−2 < 0 ∀K,L > 0

Similarly, we can apply these constructs in the theory of consumer behavior. Consider the utilityfunction U(x1, x2) = xα1x

β2 . In this utility function, x1 corresponds to the amount of good 1 consumed

and x2 corresponds to the amount of good 2 consumed. In general, utility or satisfaction is defined bythe consumption of goods. Let 0 ≤ α ≤ 1 and 0 ≤ β ≤ 1

Theorem 1.6. By an axiom of consumer behavior, a rational consumer is always non-satiated. Hence,each additional unit consumed, none less the other, will always increase the total utility. That is,MUi > 0 ∀i.

If we take the first-order derivative of the utility with respect to each good

MU1 =∂U(x1, x2)

x1= fx1 = αxα−1

1 xβ2 > 0 ∀x1, x2 > 0

MU2 =∂U(x1, x2)

x2= fx2 = βxα1x

β−12 > 0 ∀x1, x2 > 0

As with production, there is also diminishing returns in utility. The law of diminishing marginalutility states that the more of any one good is consumed in a given period, the less satisfaction (utility)is generated by consuming each additional (marginal) unit of the same good. If we have a utility functionU(x1, x2), then the assumption of diminishing marginal utility of good i is stated mathematicallyas Uxixi < 0

Theorem 1.7. The second order cross partial derivatives of the utility function should be negative bythe law of diminishing marginal utility. That is, fxixi < 0 ∀xi > 0

fx1x1 = (MU1)x1 =∂2U(x1, x2)

x21

= α(α− 1)xα−21 xβ2 < 0 ∀x1, x2 > 0

fx2x2 = (MU2)x2 =∂2U(x1, x2)

x22

= β(β − 1)xα1xβ−22 < 0 ∀x1, x2 > 0

Using the production and utility functions, prove the properties and theorems described thusfarby taking the first and second order partial derivative.

1. U(x1, x2) = 2x1x2 + 3x21 + 3x2

2

2. F (K,L) = K0.5L0.5

1.5.4 Total Differentials

Let z = f(x, y). The partial derivative fx determines the effect of x on z while holding y constant.What if both x and y are changing? The joint effect would then be obtained by taking the sum of theirindividual effects. Thus, we extend the concept of the differential to functions of two variables. Giventhe differentials dx and dy, the total differential dz is given by

dz = fxdx+ fydy

In the form above, dz approximates change in z and is due to the sum of the changes dx and dy.

19


For example, say we want to take the total differential of the function z = y2 − 4xy. To obtain thiswould be straightforward

dz = fxdx+ fydy = −4y dx+ (2y − 4x) dy

Take the total differential of the following functions

1. z = ln(xy)

2. z = 4x2 + 8xy2 + 9x2y

2 Applications of Differentiation

2.1 Taylor Approximation

Polynomial approximations of differentiable functions are useful in mathematical analysis. Taylor’sTheorem states that a differentiable function can be approximated by a polynomial. If f is an n-timesdifferentiable function, then the Taylor approximation to f about x = x0 is:

f(x) ≈ f(x0) +f ′(x0)

1!(x− x0) +

f ′′(x0)

2!(x− x0)2 + . . .+

f (n)(x0)

n!(x− x0)n (13)

We refer to the linear approximation as the first-order Taylor approximation and the quadraticapproximation as the second- order Taylor approximation.

2.1.1 Linear Approximation

If f is a differentiable function, then the linear approximation (first-order Taylor approximation) to fabout x = x0 is:

f(x) ≈ f(x0) + f ′(x0)(x− x0)

Note that when x = x0, the function f(x) and the tangent line f(x0) +f ′(x0)(x−x0) have the sameslope.

2.1.2 Quadratic Approximation

If f is a twice-differentiable function, then the quadratic approximation (second-order Taylor approxi-mation) to f about x = x0 is

f(x) ≈ f(x0) + f ′(x0)(x− x0) +1

2f ′′(x0)(x− x0)2

For example, say you were given with the function f(x) = ex. Find the first and second order Taylorpolynomial about x0 = 0.

The first-order Taylor polynomial about x0 = 0 is

P1(x) = e0 + e0(x− 0)

P1(x) = 1 + x

The first-order Taylor polynomial about x0 = 0 is

P2(x) = e0 + e0(x− 0) +1

2e0(x− 0)2

P2(x) = 1 + x+1

2x2

20


Figure 6: Illustrating the First and Second Order Taylor Polynomial

In the figure above, we see the red line corresponds to the original function f(x) while the blue linecorresponds to the linear Taylor approximation P1(x). Lastly, the green corresponds to the quadraticTaylor approximation P2(x).

Find the first-and second order Taylor polynomial for the following functions

1. f(x) = ln(1 + x) about x0 = 0

2. f(x) = x3 − 10x2 + 6 about x0 = 3

2.2 Optimization

2.2.1 Local Maxima and Minima

The function f has a local (relative) maximum value at the number c if there exists an open intervalcontaining c ,on which f is defined, such that f(c) ≥ f(x) for all x in this interval. Conversely, thefunction f has a local (relative) minimum value at the number c if there exists an open intervalcontaining c,on whichf is defined, such that f(c) ≤ f(x) for all x in this interval.

(a) Local Maximum (b) Local Minimum

Figure 7: Illustrating a Local Maxima or Minima

Theorem 2.1. If f has a local minimum or a local maximum at x∗, and if f ′(x∗) exists, then f ′(x∗) = 0.(Fermat’s Theorem)

21


Notice that at point c, whether a local minimum or maximum, the slope of the tangent line (deriva-tive) at these points are zero. More formally, the tangents at (x∗, f(x∗)) and at (x∗, g(x∗)) below arehorizontal; hence, f ′(x∗) and g′(x∗) are equal to 0.

(a) Local Minimum (b) Local Maximum

Figure 8: Illustrating Fermat’s Theorem

Note that the Fermat’s Theorem requires differentiability at the point x∗. When the derivative atx∗ does not exist, it is still possible for a function to have a local maximum or a local minimum at x∗,but the derivative will not be equal to zero at x∗.

(a) Graph of y =√|x| (b) Graph of y = 1− |x|

Figure 9: Special Cases where f ′(x) 6= 0

If x∗ is a number in the domain of function f , and if f ′(x∗) = 0, then x∗ is a critical number ofthe function and we call that point (x∗, f(x∗)) a stationary point.We can rephrase Fermat’s theoremas follows: If f has a local maximum or minimum at x∗, then x∗ is a critical number of f .

2.2.2 The First-Derivative Test

If the derivative of a function f ′(x) changes sign around a critical point x∗, the function is said to havea local (relative) optimum at that point.

• If f ′(x) changes from positive to negative, the function has a local maximum at the critical pointx∗.

• If f ′(x) changes from negative to positive, the function has a local minimum at the critical pointx∗.

• If f ′(x) does not change in sign, then the function has neither a local maximum nor local minimumat the critical point x∗.

This technique that is used to determine local maximum or minimum values is called the first-derivative test for local optima. Take note that f ′(x∗) = 0 is a necessary (first-order) conditionfor f(x∗) to be an optimal value of the function.

For example., determine the local optima of f(x) = 13x

3− 4x by using the First Derivative Test. Tostart, we find the first derivative of the function first.

f ′(x) = x2 − 4 = (x+ 2)(x− 2)

22


Setting f ′(x) = 0, we can obtain the critical values

x2 − 4 = 0

x2 = 4

x = ±2

Therefore, the critical numbers are x∗ = 2 or x∗ = −2. Since we have two critical points, we cangenerally observe three potential ranges. These are x < −2, −2 < x < 2, and x > 2. Determining atwhich critical value is a maximum/minimum lie in identifying the behavior at each point and analyzing.

If x < −2, say let’s use x = −3

f ′(x) = x2 − 4 = (−3)2 − 4 = 9− 4 = 5 > 0

Since the sign is positive, then it means that the points immediately before the point of −2 have apositive slope.

If −2 < x < 2, say let’s use x = 0

f ′(x) = x2 − 4 = (0)2 − 4 = −4 < 0

Since the sign is negative, then it means that within the range, the slope of the function is negative,which suggests that the point it is approaching to (which is x = 2) is a minimum point. To confirm thisassertion, let us investigate when x > 2.

If x > 2, say let’s use x = 3f ′(x) = x2 − 4 = (3)2 − 4 = 5 > 0

Hence, for this range, the function is positively sloped. The signs show us that f has a local maximumat x∗ = −2 and a local minimum at x∗ = 2. We graph this function for confirmation.

Figure 10: Illustrating f(x) = 13x

3 − 4x

Determine the local optima of the following functions using the First Derivative Test

1. f(x) = 15x

5 − 13x

3

2. f(x) = x3 − 3x− 4

23


2.3 Concavity and Convexity

2.3.1 Strict Concavity and Convexity

The graph of a function f is said to be strictly concave at the point (a, f(a)) if f ′(a) exists and if thereis an open interval I containing a such that for all values of x 6= a in I, the point (x, f(x)) on the graphis below the tangent line to the graph at (a, f(a)). In other words, a function f is said to be strictlyconcave on the interval [a, b] if and only if the line segment joining any two points of the graph of flies entirely below the graph. Recall that concavity (or convexity) in an interval is determined by thesign of the second derivative. Whenever f ′′(x) < 0, the graph of f for that interval is strictly concavein shape.

Conversely, the graph of a function f is said to be strictly convex at the point (a, f(a)) if f ′(a)exists and if there is an open interval I containing a such that for all values of x 6= a in I ,the point(x, f(x)) on the graph is above the tangent line to the graph at (a, f(a)). In other words, a function fis said to be strictly convex on the interval [a, b] if and only if the line segment joining any two pointsof the graph of f lies entirely above the graph. Recall that concavity (or convexity) in an interval isdetermined by the sign of the second derivative. Whenever f ′′(x) > 0, the graph of f for that intervalis strictly convex in shape.

(a) Strict Concavity (b) Strict Convexity

Figure 11: Illustrating Strict Concavity and Convexity

2.3.2 Weak Concavity and Convexity

The graph of a function f is said to be concave or weakly concave at the point (a, f(a)) if f ′(a)exists and if there is an open interval I containing a such that for all values of x 6= a in I, the point(x, f(x)) on the graph is on or below the tangent line to the graph at (a, f (a)). In other words, afunction f is said to be concave or weakly concave on the interval [a, b] if and only if the line segmentjoining any two points of the graph of f lies on or below the graph. Recall that concavity (or convexity)in an interval is determined by the sign of the second derivative. Whenever ′′(x) ≤ 0, the graph of f forthat interval is convex or weakly concave in shape.

On the other hand, the graph of a function f is said to be convex or weakly convex at the point(a, f(a)) if f ′(a) exists and if there is an open interval I containing a such that for all values of x 6= ain I, the point (x, f(x)) on the graph is on or above the tangent line to the graph at (a, f (a)). Inother words, a function f is said to be convex or weakly convex on the interval [a, b] if and only if theline segment joining any two points of the graph of f lies on or above the graph. Recall that concavity(or convexity) in an interval is determined by the sign of the second derivative. Whenever ′′(x) ≥ 0, thegraph of f for that interval is convex or weakly convex in shape.

24


(a) Weak Concavity (b) Weak Convexity

Figure 12: Illustrating Weak Concavity and Convexity

2.3.3 Slope and Shape Using the First and Second order Derivatives

The first order derivative may generally be considered to imply the slope of f(x). If f ′(x) > 0, thismeans that f(x) increases as x increases. Hence, in this case, the function has a positive slope. Con-versely, if f ′(x) < 0, this means that f(x) decreases as x increases. This implies that the function hasa negative slope. If f ′(x) = 0, then the function is at an extremum or an inflection point.

The second-order derivative implies the shape of f(x). If f ′′(x) > 0, then the slope f ′(x) isincreasing as x increases. Conversely, if f ′′(x) < 0, then the slope f ′(x) is decreasing as x increases.Lastly, if f ′′(x) = 0, then the slope f ′(x) is constant.

2.4 The Second Derivative Test

If f is twice differentiable at x∗, we can extend the implications of the first order derivative test todetermine whether the function is at a local minimum or a maximum without having to do the rigorousinterval testing.

• If f ′(x∗) = 0 and f ′′(x∗) > 0, then f has a local minimum at x∗

• If f ′(x∗) = 0 and f ′′(x∗) < 0, then f has a local maximum at x∗

• If f ′(x∗) = 0 and f ′′(x∗) = 0, then f , then the test is inconclusive

The second-order derivative test provides us a sufficient (second-order) condition for f(x∗) to bean optimal value.

(a) Showing a Maxi-mum

(b) Showing a Mini-mum

Figure 13: Illustrating the Second Derivative Test

For example, say you were tasked to determine the local optima of f(x) = x+ 1x . First, we should

get the first order derivative and equate to zero per the procedure of the first-derivative test.

f ′(x) = 1− 1

x2= 0

x = ±1

25


After which, we get the second order derivative

f ′′(x) =2

x3

We evaluate the second derivative at thee critical values. At x∗ = −1, f ′′(−1) = 2(−1)3

= −2 < 0;hence, f has a local maximum at x∗ = −1. Meanwhile, at x∗ = 1, then f ′′(1) = 2

(1)3= 2 > 0; hence, f

has a local minimum at x∗ = 1.

Figure 14: Illustrating f(x) = x+ 1x

Determine the local optima of the following functions using the First Derivative Test and SecondDerivative Test

1. f(x) = 13x

3 − 4x

2. f(x) = x3 − 3x− 4

2.4.1 Inflection Point

In some cases, the Second-Derivative Test will not work when identifying whether a critical point isa local maximum or a minimum, i.e. it will yield an inconclusive result (f ′′(x∗) = 0) even thoughthe critical point is actually a maximum or a minimum. In this case, you will have to use the FirstDerivative Test to check for maximum or minimum using the regular interval testing.

The point (a, f(a)) is a point of inflection of the graph of f if the graph has a tangent line at thatpoint where either:

• f ′′(x) < 0 if x < a and f ′′(x) > 0 if x > a; or

• f ′′(x) < 0 if x > a and f ′′(x) > 0 if x < a

In other words, a point is an inflection point if f is continuous there and the curve changes fromconcave to convex or from convex to concave at the point. At the point of inflection, if f ′′(a) exists,then f ′′(a) = 0 or f ′′(a) is undefined.

For example, say you were given f(x) = x3 − 6x2 + 9x + 1. Of course, to get the second orderderivative, we would need to get the first order derivative.

f ′(x) = 3x2 − 12x+ 9

Getting the second derivative of this function, we get

f ′′(x) = 6x− 12

26


To get the inflection point, we set this second order derivative equal to zero

0 = 6x− 12

x = 2

Therefore, when x = 2, the f ′′(x) = 0, suggesting that the function is at an inflection point. Ifx < 2, then f ′′(x) < 0, which means that for this range, the graph is concave. Conversely, when x > 2,then f ′′(x) > 0, which means that the graph is convex in this domain. To find the specific inflectionpoint, we need the coordinate value of f(x). This can be done by evaluating f(x) at the inflection xvalue computed which is x = 2.

f(2) = (2)3 − 6(2)2 + 9(2) + 1 = 8− 24 + 18 + 1 = 3

Hence, the inflection point for this function is at point (2, 3). We verify this by graphing out thefunction.

Figure 15: Illustrating f(x) = x3 − 6x2 + 9x+ 1

In general, an inflection point may look like the two illustrations below. We see that our exampleabove corresponds more to the second illustration. The inflection point "essentially" bisects the concaveand convex portions of the graph.

(a) Horizontal Tangent Line (b) Diagonal Tangent Line

Figure 16: Illustrating and Inflection Point

Note that f ′′(a) = 0 is a necessary condition for an inflection point. However, it is possible forf ′′(a) = 0 at a point a that is not an inflection point.

Find the inflection points of f(x) = x4 − 4x3 + 10

27


2.5 Global Optimum

A function f with domain D has a global (or absolute) minimum at x∗ if f(x∗) ≤ f(x) for all x in thedomain. A function f with a domain D has a global maximum at x∗ if f(x∗) ≥ f(x) for all x in thedomain.

2.5.1 Weierstrauss’ Theorem

Theorem 2.2. If f is continuous on a closed interval [a, b], then f attains a global maximum valuef(c1) and a global minimum value f(c2) at some numbers c1 and c2 within [a, b].

Figure 17: Illustrating Weierstrauss’ Theorem

We often refer to the Weirstrauss’ Theorem as the extreme value theorem. Note that the ExtremeValue Theorem fails if the interval is not closed. For example, the function f(x) = x does not havea maximum value on (0, 1). Further note that continuity of the function is required at all points ofthe interval [a, b]. Otherwise, a discontinuity could destroy the conclusion of the theorem. For exam-ple, say the function is discontinuous at x = 0. As x approaches zero from both sides, f(x) goes upto infinity. In this case, no maximum is attained. In application, when the interval is closed, we notonly examine critical values as candidates for optima, butwe also test the endpoints of the interval.

For example, say you were asked to find the global maxima and minima of f(x) = x4 − 2x2 + 1 onthe closed interval [2, 2]. We first use the first derivative test to obtain the critical values.

f ′(x) = 4x3 − 4x = 0

4x(x2 − 1) = 0

4x(x− 1)(x+ 1) = 0

Therefore, the critical values of x are x∗ = −1, x∗ = 0, and x∗ = 1. We now evaluate the functionf(x) at these critical values, as well as at the endpoints x = −2 and x = 2.

Testing the critical values of x

f(−1) = (−1)4 − 2(−1) + 1 = 0

f(1) = (1)4 − 2(1) + 1 = 0

f(0) = (0)4 − 2(0) + 1 = 1

Testing the endpoints

f(−2) = (−2)4 − 2(−2) + 1 = 9

f(2) = (2)4 − 2(2) + 1 = 9

Thus, in the interval [−2, 2], the global maxima occur at both x = −2 and x = 2, the global minimaoccur at x = −1 and x = 1, and we have a local maximum at x = 0. This conclusion is graphed in thefigure to follow.

28


Figure 18: Illustrating f(x) = x4 − 2x2 + 1 on [2, 2]

Find the global maximum and global minimum of f(x) = 13x

3 − 52x

2 + 4x− 3 on [0, 5]. Confirmyour answers by comparing it with the points on the figure below.

Figure 19: Illustrating f(x) = 13x

3 − 52x

2 + 4x− 3 on [0, 5]

2.5.2 Rolle’s Theorem

Theorem 2.3. If f is continuous on the closed interval [a, b], differentiable on the open interval (a, b)and that f(a) = f(b) is true, then there is a number c ∈ (a, b) such that f ′(c) = 0.

(a) For a Concave Case (b) For a Linear Case (c) For a Convex Case

Figure 20: Illustrating Rolle’s Theorem

29


2.6 Profit Maximization

Optimization may be used to solve for the highest profit a firm may be able to attain. Note thatthis variant is an unconstrained optimization in a univariate case, that is, no constraints and only onevariable is of interest.Consider the following example. A firm has a cost function equal to C(q) =415q

3 − 5q2 + 80q + 80 where q is the output produced. The inverse demand function for its product isp = 100− 2q, where p is price. Find the maximum profit of the firm

The profit function is defined as the difference between revenue and cost. If profit is positive, thenthe firm is earning positive returns enough to cover costs and have additional funds for reinvestment ordividends.

π(q) = R(q)− C(q) (14)

In our example, we can calculate the profit function as

π(q) = pq − C(q)

π(q) = (100− 2q)q − 4

15q3 + 5q2 − 80q − 80

π(q) = 100q − 2q2 − 4

15q3 + 5q2 − 80q − 80

π(q) = − 4

15q3 + 3q2 + 20q − 80

We then calculate f ′(q) and set it equal to zero

f ′(q) =dπ

dq= −4

5q2 + 6q + 20

Solving for q, we get the critical values q∗ = −5/2 and q∗ = 10. We discard the negative solutionsince there is no negative output. We now check if the critical value q∗ = 10 gives us a maximum. Afterthis, we solve for the second derivative of π(q)

f ′′(q) =d2π

dq2= −8

5q + 6

At q∗ = 10. we have

f ′′(10) = −8

5(1) + 6 = −10 < 0

Since f ′′(q∗) < 0, this implies a maximum profit. We can solve for the maximum profit by substi-tuting q∗ = 10 to our profit function.

π(q∗) = − 4

15(10)3 + 3(10)2 + 20(10)− 80 =

460

3

Therefore, the maximum profit is π∗ = 4603 .

Solve the following problems.

• The inverse demand function for a commodity is given by p = 41 − 0.33q and the costof producing that commodity is c(q) = −0.02q2 + 10q + 80. Find the output level thatmaximizes profit and verify this indeed yields a maximum. Find the maximum profit aswell.

• The total cost of producing x radio sets per day is given by c(q) = 14x

2 + 35x + 25 andthe price per set at which they may be sold is p(x) = 50− 1

2x. Find the output level thatmaximizes profit and verify this indeed yields a maximum. Find the maximum profit aswell.

30


3 Integration

3.1 Anti-differentiation and Indefinite Integrals

3.1.1 The Antiderivative

A function F is called an antiderivative of f on an interval I if F ′(x) = f(x) for all x in I. Theprocess of finding all antiderivatives of a function is called antidifferentiation.

Lemma 3.1. If f and g are two functions defined on an interval I, such that f ′(x) = g′(x) for all x inI, then there is a constant K such that f(x) = g(x) +K for all x in I

Theorem 3.2. If F is an antiderivative of f on an interval I, then the most general antiderivative off on I is F (x) + C, where C is an arbitrary constant.

To understand the theorem and the lemma, let us start with an example. For example, find theantiderivative of f(x) = 5. Since the derivative of 5x is 5, then F (x) = 5x. Note however, thatG(x) = 5x+ 3 is also an antiderivative of 5. Hence, the general antiderivative of f(x) = 5 is F (x) + Cor 5x+ C where C is an arbitary constant.

If F is any antiderivative of f on an interval I, then the most general antiderivative of f on I iscalled the indefinite integral and denoted as∫

If(x) = F (x) + C

The symbol∫

is called the integral sign, f(x) is the integrand, x is called the integration variable,and C is the constant of integration. The process of obtaining the indefinite integral is called integra-tion

3.1.2 Power Rule

As with differentiation, there are a couple of rules which we follow as well.∫xndx =

1

n+ 1xn+1 + C, where n is any real number (15)

For example, say you were tasked to get the integral of x4.∫x4dx =

1

5x5 + C

3.1.3 Constant Number Rule

The integral of a constant c is just cx in addition to some constant of integration∫c dx = cx+ C, where c is a constant number (16)

For example, say you were tasked to get the integral of 8.∫8dx = 8x+ C

3.1.4 Factorization and Separation Rules

When we have a form cf(x) and we want to take the derivative of this expression, we can just opt tomultiply c to the integral of f(x).∫

cf(x) dx = c

∫f(x) dx, where c is a constant number (17)

31


In terms of separability, if we were tasked to take the integral of a form like f(x)± g(x), then thiscan be done by just getting the integral of f(x) and adding (subtracting) this to the integral of g(x).∫

[f(x)± g(x)] dx =

∫f(x) dx±

∫g(x) dx (18)

For example, say you were asked to get the integral of 4x2 + 3x∫4x2 + 3x dx = 4

∫x2 dx + 3

∫x dx = 4 · 1

3x3 + 3 · 1

2x2 + C =

4

3x3 +

3

2x2 + C

3.1.5 Logarithmic and Exponential Rules

The integral of ex is just ex in addition to the constant of integration. This should be straightforwardas the derivative of x is also ex. ∫

ex dx = ex + C (19)

The integral of x−1 is specifically ln|x| in addition to some constant of integration.∫x−1 dx = ln |x|+ C (20)

The integral of ax where a is some real constant, the integral is just 1ln(a)a

x in addition to someconstant of integration. ∫

ax dx =1

ln(a)ax + C (21)

Obtain the indefinite integral of the following functions

1.∫x8 dx

2.∫

1x2

dx

3.∫

5x dx

4.∫

7x dx

5.∫

(8ex + 3x) dx

6.∫

(7ex + 9x5) dx

7.∫

(ex + xe) dx

8.∫

(4x3 − 3x2) dx

9.∫

(3x5 + 4x3/2 − 2x−1/2) dx

10.∫ (

4x10+5x7

)dx

32


3.2 Area Under a Curve and Definite Integrals

The area problem is finding the area of the region S that lies under the curve y = f(x) from a to b.This means that S is bounded by (1) the graph of a continous function, (2) the vertical lines x = a andx = b, and (3) the x-axis, The region S is called the area under the graph of f .

Figure 21: Illustrating the Area Problem

For example, let f(x) = x2 and consider the region S under the graph of f on the closed interval[0, 1]. We note that the area of S must be somewhere between 0 and 1 because S is contained in asquare of side length equal to 1. To obtain a better approximation of the area of S, we construct fournon-overlapping rectangles as follows: We divide the interval [0, 1] into four subintervals of equal lengthof 1

4 . That is, we form four rectangles, the range of which are [0, 14 ], [1

4 ,24 ], [2

4 ,34 ], [3

4 , 1].

After this, we construct the four rectangles with these subintervals as bases and with heights givenby the values of the function at either endpoints of the subintervals.

Figure 22: Using Rectangles for an Integral

Using the left endpoints of the subintervals, each rectangle has a width of 14 and heights equal to

(0)2, (14)2, (2

4)2, (34)2. If we let L4 be the sum of the areas of these approximating rectangles, we get

L4 =

(1

4

)(0) +

(1

4

)(1

16

)+

(1

4

)(4

16

)+

(1

4

)(9

16

)= 0.21875

Using the right endpoints of the subintervals, each rectangle has a width of 14 again and heights

equal to (0)2, (14)2, (2

4)2, (34)2. If we let R4 be the sum of the areas of these approximating rectangles,

we get

R4 =

(1

4

)(1

16

)+

(1

4

)(4

16

)+

(1

4

)(9

16

)+

(1

4

)(16

16

)= 0.46875

We note that the area A of S must be between L4 and R4; that is, L4 < A < R4 or 0.21875 <A < 0.46875. We obtain better estimates by increasing the number of strips.The table below shows theresult of similar calculations using n rectangles, whose heights are found with left endpoints Ln or rightendpoints Rn. A good estimate is obtained by averaging the numbers.

33


n Ln Rn

10 0.285 0.38520 0.309 0.35930 0.317 0.35050 0.323 0.343100 0.328 0.3381000 0.333 0.333

Notice that when the number of rectangles is increased (n→∞), the difference between the Ln andthe Rn nears zero. In general, the width of the interval [a, b] is b− a, so the equal width of each of then rectangles is

∆x =b− an

The bases of the rectangles divide the interval into n subintervals, [a, x1], [x1, x2], [x2, x3], ..., [xn−1, b].Instead of using the left or right endpoints, we could take the height of the ith rectangle to be the valueof f at any number x∗i in the ith subinterval [xi−1, xi]. We call the numbers x∗1, x∗2, ..., x∗n the samplepoints.

3.2.1 Reimann Sum

The area A of the region S that lies under the graph of the continuous function f is the limit of thesum of the areas of approximating rectangles.

A = limn→∞

n∑i=1

f(x∗i )∆x (22)

We call the sum on the right hand side of the expression as the Reimann sum. It approximatesthe area under the curve from a to b. This approximation gets better and better as n increases.

3.2.2 Definite Integral

If f is a continuous function defined for x ∈ [a, b], the definite integral of f from a to b is equal toequation 22 if the limit exists ∫ b

af(x) dx = lim

n→∞

n∑i=1

f(x∗i )∆x (23)

When they say f is a Riemann-integrable or integrable, a and b are called the limits of integration.a is the lower limit and b is the upper limit.

Lemma 3.3. If f is continuous, the limit in the definition always exists and gives the same value nomatter how we choose the sample points. This implies that a function is integrable only if it is continuous.Therefore, continuity is a necessary condition for integrability

3.2.3 Fundamental Theorems of Calculus

Let the function f be continous on the closed interval [a, b] and let t be any number in [a, b]. If g is thefunction defined by

g(t) =

∫ t

af(x) dx

Then it is reasonable to conclude that g′(t) = f(t). Note that it should follow that g is also acontinuous function on [a, b].

34


Theorem 3.4. Fundamental Theorem of Calculus 1: The derivative of an anti-derivative (i.e. theintegral) is the original function itself

If f is continous on [a, b], then the expression below were F is any antiderivative of f is the definiteintegral. ∫ b

af(x) dx = F (b)− F (a)

Theorem 3.5. If the antiderivative of the function being integrated is known, then the integral may beevaluated by subtracting the values of the antiderivative at the endpoints of the interval.

We often use the notation below to denote this operation

F (b)− F (a) = F (x)|baFor example, say you were asked to solve the following integral.∫ 2

1(3x2 − 4x) dx

We first apply the factorization techniques

3

∫ 2

1x2 dx− 4

∫ 2

1x dx

We apply the rules of integration discussed previously and follow the operation of the definite integral

3

(1

3x3

∣∣∣∣21

)− 4

(1

2x2

∣∣∣∣21

)Substitute the limits of integration using the order F (b)− F (a).

3

((2)3

3− (1)3

3

)− 4

((2)2

2− (1)2

2

)Simplifying

3

(7

3

)− 4

(3

2

)= 7− 6 = 1

Therefore, the solution to our problem is 1 which is also under the area of the curve f(x) = 3x2−4xfrom the interval [1, 2]. Notice that we have no need for our constant/coefficient of integration. This isbecause we have a starting and ending limit represented by the bounds or limits of integration.

Evaluate the following definite integrals

1.∫ 2−2 x

3 dx

2.∫ 1−1 e

x dx

3.∫ 1

0 (x−√x) dx

4.∫ 1

0 (4 + 3x2) dx

5.∫ 4

0

√x dx

6.∫ 3

1 ln(x) dx

7.∫ 2

4

(u2

3 + 1)

du

35


Figure 23: Showing a Shaded A of f(x) = x2 − 1

For example, let us graph part of the function f(x) = x2 − 1

Suppose you were asked to determine the shaded area in the graph. We can get this using a definiteintegral. Note that f(x) < 0 on [0, 1]. The (definite) integral of f from 0 to 1 is the shaded area.∫ 1

0(x2 − 1) dx =

∫ 1

0

(1

3x3 − x

)=

1

3− 1 = −2

3

The Fundamental Theorem of Calculus is not a procedure for finding indefinite integrals, but forevaluating definite integrals, given that we known an indefinite integral of the integrand.

3.2.4 Additional Rules on Definite Integrals

The rules from definite integration are largely the same as indefinite integration. However, there are afew additions or modifications to the existing rules. First, the constant function rule can be extendedto accommodate the bounds ∫ b

ac dx = c(b− a) (24)

Taking the negative of the integral merely gives you the integral with the upper and lower boundsreversed ∫ b

af(x) dx = −

∫ a

bf(x) dx (25)

Say there was a point c which is inside [a, b]. Then, the definite integral may be split into a simplesum for as long as the ends add up.∫ b

af(x) dx =

∫ c

af(x) dx +

∫ b

vf(x) dx (26)

If f(x) ≥ 0 for x ∈ [a, b], then ∫ b

af(x) dx ≥ 0

If f(x) ≥ g(x) for x ∈ [a, b], then ∫ b

af(x) dx ≥

∫ b

ag(x) dx

If m ≤ f(x) ≤M for x ∈ [a, b], then the form below holds where m and M are real numbers.

m(b− a) ≤∫ b

af(x) dx ≤M(b− a)

36


3.3 Integration by Substitution

If u = g(x) is a differentiable function whose range is an interval I and f is continuous on I, then∫f(g(x))g′(x) dx =

∫f(u) du (27)

Notice that if u = g(x), then du = g′(x) dx. Integration by substitution is used to evaluate integralswhich cannot fully be solved using just the typical rules or techniques.

1. Let u = g(x), where g(x) is part of the integrand, usually the "inside function" of the compositefunction.

2. Find du = g′(x) dx

3. Use the substitution u = g(x) and du = g′(x) dx to convert the entire integral into one involvingonly u.

4. Evaluate the resulting integral

5. Replace u by g(x) to obtain the final solution as a function of x.

For example, say we evaluate∫

(x2 + 1)52x dx.

Let u = x2 + 1 or the "inner function". Then du = 2x dx. Substituting into the integral, we get∫(x2 + 1)52x dx =

∫u5 du

We can now apply the basic rules of integration∫u5 du =

u6

6+ C

Substituting the initial function back

(x2 + 1)6

6+ C

Evaluate the following integrals

1.∫

(x2 + 50x+ 9)20(2x+ 50) dx

2.∫

2(2x+ 4)5 dx

3.∫ex

2+1 dx

4.∫ekx dx, k is a constant

5.∫ 4

0

√2x+ 1 dx

6.∫ 2

0 (y2 + 1)32y dy

37


3.4 Integration by Parts

3.4.1 Integration by Parts of Indefinite Integrals

If f(x) and g(x) are differentiable functions that are integrable over [a, b], then∫f(x)g′(x) dx = f(x)g(x)−

∫g(x)f ′(x) dx (28)

If we let u = f(x) and v = g(x), then du = f ′(x) dx and dv = g′(x) dx. We can state the resultabove into the more conventional and well known integration by parts form.∫

u dv = uv −∫v du (29)

The success of using integration by parts depends on the proper choice of u and dv. In general,choose u and dv such that

• du is simpler that u, and

• dv is easy to integrate

For example, suppose we evaluate ∫xex dx

We let u = x and dv= ex dx. This means that du=dx and v =∫ex dx = ex. Using the integration

by parts formula, we get ∫u dv = uv −

∫v du∫

xex dx = xex −∫ex dx

xex − ex + C

Evaluate the following indefinite integrals

1.∫xe2x dx

2.∫x2 ln(x) dx

3.∫xex dx

4.∫x√x dx

3.4.2 Integration by Parts for Definite Integrals

If f(x) and g(x) are differentiable functions that are integrable over [a, b], then∫ b

af(x)g′(x) dx = [f(b)g(b)− f(a)g(a)]−

∫ b

ag(x)f ′(x) dx

Let u = f(x) and v = g(x), then du= f ′(x) dx and dv= g′(x) dx, we can state the result as∫ b

au dv = uv|ba −

∫ b

av du (30)

Alternatively, you can solve the indefinite integral first and apply the limits of integration afterwards.For example, recall our previous derivation for the integral of xex

38


∫xex dx = xex −

∫ex dx = xex − ex

Applying the rules of definite integration

ex(x− 1)

e2(2− 1)− e1(1− 1) = e2 ≈ 7.39

3.5 Integration by Using Partial Fractions

Consider the integral ∫2x+ 1

x2 − 2x− 3dx

When the integrand is a rational function like the one above, integration may be performed by firstdecomposing it into the sum of simpler rational functions called partial fractions. These fractionscan then be integrated using typical methods. The method of decomposing a rational function isstraightforward when the degree of the numerator is less than the degree of the denominator.

For example, let us use the integral above which is∫

2x+1x2−2x−3

dx. To evaluate this integral, we firsthave to factor the denominator

x2 − 2x− 3 = (x− 3)(x+ 1)

We then set the original function be a sum of two partial fractions

2x+ 10

x2 − 2x− 3=

A

x− 3+

B

x+ 1

Multiplying both sizes by x2 − 2x− 3 to cancel out all the denominators. Note that x2 − 2x− 3 =(x− 3)(x+ 1)

2x+ 10 = A(x+ 1) +B(x− 3)

We then look for values of x that will eliminate the terms involving A and then B. This will enableus to solve for the unknowns A and B. In this case, we have x = −1 to eliminate A and x = 3 whichwill eliminate B.

When x = −1, we get

−2 + 10 = A(0) +B(−4); hence, B = −2

When x = 3, we get

6 + 10 = A(4) +B(0); hence, A = 4

Therefore

2x+ 10

x2 − 2x− 3=

4

x− 3+−2

x+ 1

We can now perform the integration∫2x+ 10

x2 − 2x− 3dx =

∫4

x− 3dx +

∫−2

x+ 1dx

4

∫1

x− 3dx− 2

∫1

x+ 1dx

4 ln(x− 3)− 2 ln(x+ 1) + C

39


Evaluate the following indefinite integrals

1.∫

x+2x2−2x

dx

2.∫

3x2−25

dx

3.∫

x2+4x3−3x2+2x

dx

3.6 Integration by Using Partial Fractions with Repeated Factors

A rational function whose denominator has repeated factors (x−a)k must be decomposed into the sumof k partial fractions, one for each power of (x − a). After decomposing, the fractions can then beintegrated using typical methods. Again, this method of decomposing a rational function is easy to usewhen the degree of the numerator is less than the degree of the denominator.

For example, say you were given the integral∫3x+ 5

x3 − 2x2 + xdx

We can rewrite the integrand as

3x+ 5

x3 − 2x2 + x=

3x+ 5

x(x− 1)2=A

x+

B

x− 1+

C

(x− 1)2

Multiplying both sides of the equation by x(x− 1)2, we get

3x+ 5 = A(x− 1)2 +Bx(x− 1) + C(x)

We now find values of x that can cancel out some of the letters. We have x = 1 which will eliminateA and B and x = 0 which will eliminate B and C.

When x = 1, we get

3 + 5 = A(0) +B(0) + C(1); hence C = 8

When x = 0, we get

5 = A(1) +B(0) + C(0); hence A = 5

Since A and C are already known, we may substitute any number to solve for B, say x = 2.

3(2) + 5 = 5(1) +B(2)(1) + 8(2); hence B = −5

Therefore

3x+ 5

x3 − 2x2 + x=

3x+ 5

x(x− 1)2=

5

x+−5

x− 1+

8

(x− 1)2

We can now perform the integration∫3x+ 5

x3 − 2x2 + xdx =

∫5

xdx +

∫−5

x− 1dx +

∫8

(x− 1)2dx

5 ln(x)− 5 ln(x− 1)− 8

x− 1+ C

40


3.7 Improper Integrals

Integrals of the form below are called improper integrals∫ ∞a

f(x) dx

∫ b

−∞f(x) dx

∫ +∞

−∞f(x) dx

Another type of improper integral is one where the limits of integration a and b are finite but f hasan infinite discontinuity on the interval [a, b]. For example, the integrals below are improper becausefor the first, the integrand goes to infinity as x approaches zero, an interior point of [−1, 1], while forthe second, the integrand goes to infinity as x approaches zero, the lower limit of integration.∫ 1

−1

1

xdx

∫ 1

0

1

x2dx

Theorem 3.6. If∫ ta f(x) dx exists for ever number t ≥ a, then

∫∞a f(x) dx = limt→∞

∫ ta f(x) dx

provided this limit exists

The theorem basically states that an solution exists for an improper integral with an upper limit ofinfinity for as long as the limit exists.

Theorem 3.7. If∫ bt f(x) dx exists for ever number t ≤ b, then

∫ b−∞ f(x) dx = limt→−∞

∫ bt f(x) dx

provided this limit exists

The theorem basically states that an solution exists for an improper integral with an lower limit ofinfinity for as long as the limit exists.

The improper integrals∫∞a f(x) dx and

∫ b−∞ f(x) dx are called convergent if the corresponding

limit exists and divergent if the limit does not exist.

If both∫∞a f(x) dx and

∫ b−∞ f(x) dx are convergent, then we define∫ ∞−∞

f(x) dx =

∫ a

−∞f(x) dx +

∫ ∞a

f(x) dx

Figure 24: Showing∫∞−∞ f(x) dx

41


For example, determine the convergence or divergence of the following integrals.

Say we have∫∞

11x dx∫ ∞

1

1

xdx = lim

t→∞

∫ t

1

1

xdx = lim

t→∞ln(x)|t1 = lim

t→∞[ln(t)− ln(1)] =∞

This given improper integral is divergent because the limit does not exist.

Say we have∫ 0−∞ e

x dx∫ 0

−∞ex dx = lim

t→−∞

∫ 0

tex dx = lim

t→−∞ex|0t = lim

t→−∞[e0 − et] = 1

This given improper integral is convergent because the limit exists.

Say we were given with∫∞−∞ 2xe−x

2 dx. By definition, we have∫ ∞−∞

2xe−x2dx =

∫ 0

−∞2xe−x

2+

∫ ∞0

2xe−x2dx dx

We evaluate the integrals on the right-hand side term by term. To do so, we have to use integrationby substitution. Let u = −x2 and du = −2x dx

∫ 0

−∞2xe−x

2dx = lim

t→−∞

∫ 0

t2xe−x

2dx = lim

t→−∞

∫ 0

t−eu du = lim

t→−∞−e−x2 |0t = −e0 − (−e−t2) = −1

∫ ∞0

2xe−x2dx = lim

t→∞

∫ t

02xe−x

2dx = lim

t→∞

∫ t

0−eu du = lim

t→∞−e−x2 |t0 = −e−t2 − (−e0) = 1

Thus,∫∞−∞ 2xe−x

2 dx = −1+1 = 0, which means that the given integral is convergent. The reasonthe integral is zero is that the integral from negative infinity to zero is a negative area while the integralfrom zero to positive infinity is a positive area. Those areas are equal as seen below.

Figure 25: Showing 2xe−x2

Determine the convergence or divergence of the following improper integrals

1.∫∞

11x2

dx

2.∫∞

0 e−x dx

42


Theorem 3.8. If f is continuous on (a, b] and is discontinuous at a, then∫ ba f(x) dx = limt→a

∫ bt f(x) dx

provided that the limit exists

The theorem just states that if the lower limit of integration is a point of discontinuity, the integralmay still be evaluated for as long as the limit exists.

Theorem 3.9. If f is continuous on (a, b] and is discontinuous at b, then∫ ba f(x) dx = limt→b

∫ ta f(x) dx

provided that the limit exists

The theorem just states that if the upper limit of integration is a point of discontinuity, the integralmay still be evaluated for as long as the limit exists.

The improper integral∫ ba f(x) dx whether discontinuous at a or b is called convergent if the

corresponding limit exists and divergent if the limit does not exist.

If f has a discontinuity at c, where a < c < b, then we define∫ b

af(x) dx =

∫ c

af(x) dx +

∫ b

cf(x) dx

Determine the convergence or divergence of the following improper integrals

1.∫ 1

01√xdx

2.∫ 3

01√3−x dx

3.∫ 3−2

1x3

dx

3.8 Economic Applications of Integrations

3.8.1 Finding Total Cost Functions from Marginal Cost Functions

Recall that in Economics, the derivative F ′(x) of a function F (x) is called a marginal function. F (x)is then called a total function Thus, a total function is simply the integral of a marginal function.

Given a total cost function C(q), the process of differentiation can yield the marginal cost functionC ′(q). Because integration is merely the inverse operation of differentiation, integrating the marginalcost function C ′(q) should yield the total cost function C(q).

For example, the marginal cost function of a firm is given by MC(q) = 3q2 − 30q + 80. The fixedcost is known to be 100. Find the total cost function. To do this, we merely integrate the marginal costfunction and add the fixed cost.

C(q) =

∫(3q2 − 30q + 80) dq = 3

∫q2 dq− 30

∫q dq + 80

∫dq = q3 − 15q2 + 80q + 100

Hence, the total cost function is q3 − 15q2 + 80q + 100

3.8.2 Finding Total Revenue Functions from Marginal Revenue Functions

By integrating the marginal revenue function R′(q), we can find the total revenue function R(q).But total revenue R is also defined by the equation R = pq, where p is the price per unit and q is thetotal number of output produced. Thus, we have p = R/q or p(q) = R(q)/q. This means that we canget the demand function p(q) once we have the total revenue function R(q). Because it is generally truethat the total revenue is zero when q = 0, this fact can be used to evaluate the arbitrary constant when

43


finding the total revenue function from the marginal revenue function.

For example, say the marginal revenue function is given by MR(q) = 8− 6q − 2q2 where q denotesoutput produced. Find the total revenue function R. To get R, we just simply integrate the marginalrevenue function.

R(q) =

∫(8− 6q − 2q2) dq = 9q − 3q2 − 2

3q3 + C

Because R(0) = 0, we get C = 0. Hence, the total revenue function is

R(q) = 8q − 3q2 − 2

3q3

3.8.3 National Income, Consumption, and Savings

Let C(Y ) denote an economy’s consumption function where C is the household consumption and Y isthe national income. Then the marginal propensity to consume (MPC) is given by C ′(Y ), andthe consumption function C(Y ) is just the integral of the MPC with respect to Y . In addition, sinceprivate savings S is defined by the equation S(Y ) = Y − C(Y ), then the marginal propensity tosave (MPS) is given by S′(Y ) = 1− C ′(Y ).

For example, suppose the marginal propensity to save is given by the function S′(Y ) = 0.3−0.1Y −0.5,and if private savings S is zero when national income Y is Php 81, then

S(Y ) =

∫(0.3− 0.1Y −0.5) dY = 0.3Y − 0.2Y 0.5 + k

The specific value of k can be found from the fact that S = 0 when Y = 81. Substitution of thisinformation into the preceding integral will yield the value for the arbitrary constant k.

Thus we have

0 = 0.3(81)− 0.2(81)0.5 + k

k = −22.5

Therefore, the private savings function is S(Y ) = 0.3Y − 0.2Y 0.5 − 22.5 while the consumptionfunction is obtained from

S(Y ) = Y − C(Y )

0.3Y − 0.2Y 0.5 − 22.5 = Y − C(Y )

C(Y ) = 0.7Y + 0.2Y 0.5 + 22.5

3.8.4 Investment, Capital Formation, and Capital Accumulation

The process of adding to a given stock of capital is referred to as capital formation. If this processis considered to be continuous over time, capital stock K can be expressed as a function of time K(t),and the rate of capital formation is then given by dK

dt = K ′(t). The rate of capital formation at time tis the same as the rate of net investment flow at time t, denoted by I(t).

K(t) =

∫K ′(t) dt =

∫I(t) dt

In the form above, an initial condition K(0) must be specified to evaluate the arbitrary constant ofintegration.

Suppose that the net investment flow is described by I(t) = 3√t and that the initial capital stock

at time t = 0 is K(0) = K0. By integrating I(t) with respect to t, we obtain

44


K(t) =

∫3√t dt = 2t1.5 + C

Next, letting t = 0, we find K(0) = C or K0 = C. Therefore, the time path of K is

K(t) = 2t1.5 +K0

Sometimes, the concept of gross investment is used together with net investment in a model. Denot-ing gross investment by Ig and net investment by I, we can relate them to each other by the equationIg = I + δK, where δ represents the rate of depreciation of capital and δK is the rate of replacementinvestment

Since K(t) is the antiderivative or integral of I(t), we may write the definite integral∫ b

aI(t) dt = K(b)−K(a)

This indicates the total capital accumulation during the time interval [a, b]. Of course, this alsorepresents the are under the I(t) curve.

Figure 26: Showing Capital Accumulation

For example, if net investments, in thousands per year, is a non-constant flow that is determined byI(t) = 3

√t, then the capital accumulated during the time interval [0, 4] lies in the definite interval.∫ 4

03√t dt = (2t1.5)|40

(2)(4)1.5 − (2)(1)1.5 = 16− 2 = 14

We may express the amount of capital accumulation during the time interval [0, t], for any investmentrate I(t), by the definite integral

K(t)−K(0) =

∫ t

0I(t) dt

K(t) = K(0) +

∫ t

0I(t) dt

The preceding equation yields the following expression for the time path K(t). That is, the amountof K at any time t is the initial capital plus the total capital accumulation that has occurred since.

For example, suppose that a firm begins at time t = 0 with capital stock of K(0) = 500, 000and, in addition to replacing any depreciated capital, is planning to invest in new capital at the rateI(t) = 600t2. The planned level of capital stock t years from now is computed according to

K(t) = K(0) +

∫ t

0I(t) dt

45


K(t) = 500, 000 +

∫ t

0600t2 dt

If t = 10, then capital stock 10 years from now is K(10) = 700, 000.

3.8.5 Consumer, Producer, and Social Surpluses

The Consumer’s Surplus (CS) represents the welfare of consumers. It is the extra amount consumerswould have been willing to pay but did not. Thus, CS is calculated as a definite integral:

CS =

∫ Q∗

0D(Q) dQ− p∗Q∗ (31)

In the equation above, D(Q) is the inverse market demand function, p∗ is the market equilibriumprice, and Q∗ is the market equilibrium output.

On the other hand, the Producer’s Surplus (PS) represents the welfare of producers or sellers. It isthe extra amountvproducers received above what they were willing to accept. Thus, PS is calculatedas a definite integral:

PS = p∗Q∗ −∫ Q∗

0S(Q) dQ (32)

In the equation above, S(Q) is the market supply function, p∗ is the market equilibrium price, andQ∗ is the market equilibrium output.

Lastly, the Social Surplus (SS) (or economic surplus or net market surplus) is the total welfare ofboth consumer and producer. It is the sum of both consumer’s and producer’s surplus; that is CS+PSor

SS = CS + PS =

∫ Q∗

0D(Q) dQ− p∗Q∗ +

(p∗Q∗ −

∫ Q∗

0S(Q) dQ

)Simplifying, we get

SS =

∫ Q∗

0D(Q) dQ−

∫ Q∗

0S(Q) dQ =

∫ Q∗

0[D(Q)− S(Q)] dQ (33)

(a) Illustrating Consumer Surplus (b) Illustrating Producer Surplus (c) Illustrating Social Surplus

Figure 27: Illustrating the Economic Surpluses

For example, given the respective inverse market demand and supply functions P = 50− 0.1Q andP = 0.2Q+ 20, find the consumer’s, producer’s, and social surpluses. To start off, we need to computefor the equilibrium quantity and price. We know that at equilibrium, supply is equal to demand. Weneed to equate the two functions we were given with and compute for Q.

46


50− 0.1Q = 0.2Q+ 20

30 = 0.3Q

Q∗ = 100

Knowing that Q∗ = 100, we can plug this into either the inverse demand or inverse supply to getthe market equilibrium price. Say we plug it into the inverse demand function

P = 50− 0.1(100) = 40

P ∗ = 40

We now know that P ∗ = 40 and Q∗ = 100, we can now use these information in computing for theeconomic surpluses. We start with the consumer surplus.

CS =

∫ Q∗

0D(Q) dQ− p∗Q∗ =

∫ 100

0(50− 0.1Q) dq− (100)(40)

50

∫ 100

0dq− 0.1

∫ 100

0Q dq− 4000

50(Q)|1000 − 0.1

(Q2

2

)∣∣∣∣100

0

− 4000

50(100− 0)− 0.1

(1002

2− 02

2

)− 4000

5000− 500− 4000

CS = 500

Using the same information to obtain the producer surplus

PS = p∗Q∗ −∫ Q∗

0S(Q) dQ = (100)(40)−

∫ 100

0(20 + 0.2Q) dq

4000− 20

∫ 100

0dq− 0.2

∫ 100

0Q dq

4000− 20(Q)|1000 − 0.2

(Q2

2

)∣∣∣∣100

0

4000− 20(100− 0)− 0.2

(1002

2− 02

2

)4000− 1000− 2000

PS = 1000

Adding both the consumer and producer surpluses yields us the social surplus which is SS =1000 + 500 = 1500. Alternatively, as an exercise, you can verify using the formula above.

Find the equilibrium point of the following demand and supply functions and calculate the socialsurplus at equilibrium

p(q) = D(q) = − q2

100− x+ 50

p(q) = S(q) =q2

25+ 10

47


4 Matrix Algebra

4.1 Introduction to Matrices

Partial equilibrium analysis can be relatively tedious to solve via the methods of elimination or substi-tution, especially when the system contains many equations. Fortunately, through the use of matrices,we can simplify the process of computation and even expand it up to n number of goods.

A matrix is a rectangular array A of numbers (or symbols representing numbers) and variables andis written in the form

A =

a11 a12 . . . a1n

a21 a22 . . . a2n...

.... . .

...am1 am2 . . . amn

The numbers in the array A are called entries. Each entry has double subscripts which indicate the

address of the entry, i.e. aij is located in the ith row and jth column of A. A matrix with m rows andn columns is referred to as an (read m by n) matrix and m and n are called the dimensions of thematrix. If m = n, then the matrix is called a square matrix of order n.

4.1.1 Special Forms of Matrices

A vector is a matrix with either one row, called a row matrix, or one column, called a column matrix

x =(a11 a12 a13 . . . a1n

)

y =

a11

a21

a31...

am1

A scalar is a constant. In matrix formulation, this is a matrix with only one row and one column.

A =(a11

); or simplyA = (a)

A square matrix is a matrix that has the same number of rows and columns (i.e. m = n)

A diagonal matrix is a square matrix having non-zero elements only in the principal diagonal, i.e.,the diagonal running from the upper left to the lower right.a11 0 0

0 a22 00 0 a33

Note: All diagonal matrices are square matrices, but not all square matrices are diagonal matrices.

An Identity Matrix (or unit matrix ) is a diagonal matrix that has values of 1 in its principaldiagonal. Since it is a diagonal matrix, all non-diagonal entries must be zero.1 0 0

0 1 00 0 1

Note: A matrix pre or post multiplied by a conformable identity matrix is equal to the matrix itself,

i.e. an identity matrix plays the same roles as 1 in scalar algebra.A Null Matrix is a matrix that has 0’s for its entries; plays a similar role to 0 in scalar algebra.

48


0 0 00 0 00 0 0

An Idempotent Matrix is when say a matrix A multiplied to itself yields itself. That is, A ·A = A.

The simplest examples of idempotent matrices are the identity matrix and the null matrix.

A Triangular Matrix is a matrix with non-zero elements in either the northeast or southwest ofthe principal diagonal while the rest of the elements are zero. We generally subclassify these into two.

• Upper Triangular Matrix - a matrix for which every diagonal entry in the northeast, including theprincipal diagonal, is non-zero a11 a12 a13

0 a22 a23

0 0 a33

• Lower Triangular Matrix - a matrix for which every diagonal entry in the southwest, including

the principal diagonal, is non-zero a11 0 0a21 a22 0a31 a32 a33

A Symmetric Matrix is a square matrix for which the off-diagonal elements in the northeastare a mirror image of the off-diagonal elements in the southwest. 1 1 −1

1 2 0−1 0 5

There are Equal Matrices which are matrices with the same dimensions and the same corre-sponding entries in corresponding positions. 1 1 −1

1 2 0−1 0 5

=

1 1 −11 2 0−1 0 5

A Scalar Matrix is a diagonal matrix where all the entris in the principal diagonal re equal. Theidentity matrix is an example of a scalar matrix.2 0 0

0 2 00 0 2

The Transpose Matrix is the matrix which contains the entries for which the rows and columnsof the matrix are interchanged. We denote the transpose of matrix A as A′ or AT

A =

1 42 53 6

→ AT =

(1 2 34 5 6

)

There are a couple of properties that transpose matrices have

1. If A is an m× n matrix, then AT is an n×m matrix

49


2. (AT )T = A The transpose of a transpose is the original matrix3. (A±B)T = AT ±BT

4. (cA)T = cAT

5. (AB)T = BTAT

6. If A is symmetric, then A = AT

4.2 Matrix Operations

4.2.1 Matrix Addition and Subtraction

Matrix addition and subtraction requires matrices to be conformable, i.e. matrices must have the samedimensions. For example, say Matrix A has a dimension of 2×3 and Matrix B has a dimension of 2×3.We can add and subtract matrix A and B since both matrices have the same dimension. Say we havea Matrix F which has a dimension of 3× 3, we cannot add and subtract this to A or B as they are ofdifferent dimensions.

If matrices A and B are conformable, then matrix C = A+B will also have the same dimension asA and B. The entries in matrix C are obtained by simply adding (subtracting) the entries of A and B

with the same address. For example, say A =

(2 5 03 −1 8

); B =

(6 1 27 9 0

)

A+B =

(2 5 03 −1 8

)+

(6 1 27 9 0

)=

(8 6 210 8 0

)4.2.2 Scalar Multiplication

Scalar multiplication involves taking the product of a matrix and a scalar. To do so, simply multiply

each entry of the matrix by the scalar. For example, say q is the scalar and A =

(a bc d

), then:

qA =

(qa qbqc qd

)4.2.3 Matrix Multiplication

Matrix multiplication is not as simple as scalar multiplication since (1) it requires conformity be-tween the two matrices and (2) involves calculation of the matrices’ inner product.

For A · B = C, A is called the lead matrix, and B is called the lag matrix. You call this as "pre-multiplying matrix B by matrix A" or "post-multiplying matrix A by matrix B".

Theorem 4.1. The number of columns in the lead matrix must be equal to the number of rows in thelag matrix. (Conformity Condition)

For example, say matrix A is an (m×n) matrix, matrix B is an (n×k). By the conformity condition,A · B is possible since A has n columns and B has n rows. By contrast, B · A is not possible since itdoesn’t meet the conformity condition.

Dimensions of the product of two matrices would be the "outer product" AB = C where C hasthe same number of rows as the lead matrix and the same number of columns as the lag matrix. Toillustrate, each element in C has an inner product of two vectors in matrices A and B. To illustrate theconcept of the inner product, consider vectors u and v below

u =(u1 . . . un

); v =

v1...vn

50


Note that u · v satisfies the conformity condition. The inner product of these two are given by thescalar [uv] = [u1v1 + . . .+ unvn] For example, take the case below

u · v =(1 2 3

)·

51015

= [1 · 5 + 2 · 10 + 3 · 15] = [5 + 20 + 45] = [70]

In general, for A ·B = C, we havea11 . . . a1n...

. . . . . .am1 . . . amn

·b11 . . . b1k

.... . . . . .

an1 . . . ank

=

c11 . . . c1k...

. . . . . .cm1 . . . cmk

Note that A is m× n, B is n× k, and as a result of the multiplication process, C is m× k.

cij is the entry of matrix C that appears in the ith row and jth column. It is also the inner productof the ith row of A and the jth column of B.(

a11 a12

a21 a22

)·(b11

b21

)=

(c11

c21

)Note that

c11 = a11 · b11 + a12 · b21

c21 = a21 · b11 + a22 · b21

For example, say A =

(1 23 4

)and B =

(1020

). Perform A ·B(

1 23 4

)·(

1020

)=

(1 · 10 + 2 · 203 · 10 + 4 · 20

)=

(50110

)

Perform the following indicated matrix operations

1. A =

1 32 84 0

; b =

(59

), A · b

2. A =

6 3 11 4 −24 −1 5

, x =

x1

x2

x3

, A · x

3. M =

3 2 1 05 6 2 77 6 5 21 2 3 5

, U =

5 61 35 62 3

, M ·

4. Consider three consumers (i = 1, 2, 3) who purchased four goods (j = 1, 2, 3, 4). Let aijdenote the quantity of good j purchased by consumer i at price pj . The purchases may bearranged in a matrix A and the prices in a vector p

A =

a11 a12 a13 a14

a21 a22 a23 a24

a31 a32 a33 a34

a41 a42 a43 a44

, p =

p1

p2

p3

p4

What is Ap equal to? What does each entry of Ap represent?

51


While matrices, like numbers, can undergo the operations of addition, subtraction, and multiplication– subject to the conformability conditions – it is not possible to divide one matrix by another. That is,we cannot write A/B. However, instead of writing A/B, we can instead write and solve for AB−1 whereB−1 is the inverse of matrixB. But take note that the product of AB−1 may not be necessarily thesame as the product of B−1A. Therefore, the expression A/B cannot be used without causing confusionand must, thus, be avoided.

4.2.4 Properties of Matrix Operations

1. Commutative. Matrix addition is commutative, A + B = B + A, for as along as A and B areconformable for addition. Scalar multiplication is also commutative, that is, kA = Ak, where k issome scalar. Matrix multiplication is not commutative, that is, A ·B 6= B · A. Even if A ·B andB ·A are possible, A ·B = B ·A is not necessarily true.

2. Associative. Matrix addition is associative (A + B) + C = A + (B + C), for as long as A andB are conformable for addition. Matrix multiplication is associative (A ·B) · C = A · (B · C) , aslong as the conformability condition is satisfied by each pair of adjacent matrices.

3. Distributive. Matrix multiplication is distributive A · (B + C) = A ·B +A · C

4. Given scalars c, k, and matrices A,B

• (c+ k)A = cA+ kA

• (ck)A = c(kA) = k(cA)

• c(AB) = (cA)B = A(cB)

• c(A+B) = cA+ cB

4.3 Solving Systems of Linear Equations using Matrices

4.3.1 Matrix Algebra and the Matrix Form

Matrix algebra provides a compact way of writing an equation system. Once the system has beenwritten down, matrix algebra also provides a mechanism for testing the existence of a solution to anequation system. It is an alternative procedure for finding the solution to any linear equation system.Note that matrix algebra applies to linear equations only.

Matrices allow us to write equations in a compact format. The process involves identifying theendogenous variables, coefficients, and constants (along with exogenous variables), and organizing theseinto separate matrices. The equation system with m equations and endogenous variables in expandedform may be written as

a11x1 + a12x2 + . . .+ a1mxm = d1

a21x1 + a22x2 + . . .+ a2mxm = d2

...

am1x1 + am2x2 + . . .+ ammxm = dm

Given this system of equations

1. You have m equations and m endogenous variables, i.e. x1, x2, ..., xm

2. On the left hand side, you have m×m aij ’s which are the coefficients of the endogenous variables.

3. On the right hand side, you have m× 1 d’s, which are the exogenous variables or constant terms.

52


Thus, given this system of equations, we can make a matrix of coefficients, endogenous variables,and exogenous variables such that

Ax = d (34)

In this equation, we have A which is an m×m, x which is an m×1 and d which is an m×1 matrix.Putting these together we have a11 . . . a1m

.... . . . . .

am1 . . . amm

x1

...xm

=

d1...dm

This system of equations written in matrix form can now be solved to obtain the solution values of

the endogenous variables.

Rewrite the following systems of linear equations into matrix form

1.x− 2y + 3z = 7

2x+ y + z = 4

−3x+ 2y − 2z = −10

2.Qd = 500− 50P

Qs = 50 + 25P

Qd = Qs

3.C = 260 + 0.4Y

I = 300 + 0.5Y

Y = C + I + 110

4.3.2 Defining Determinants

For a square matrix, one can capture important information about the matrix in just a single numbercalled the determinant. The determinant is useful in solving systems of linear equations.Necessaryand sufficient conditions for a determinant to exist

• Necessary Condition. The matrix must be square, i.e. when transforming systems of linearequations into matrix form, the matrix will only be square if and only if the number of equationsis equal to the number of unknowns.

• Sufficient Condition. All rows and columns of a square matrix must be linearly independentof each other. Linear independence prevents systems of equations from reducing into a singleequation and, thus, ensures that we get a unique solution.

To illustrate linear independence

A =

a11 . . . a1m...

. . . . . .am1 . . . amm

=

v1...vm

In the formulation above, vi is a row vector containing the elements of row i = (ai1 . . . aim). Two

rows (vi and vj) are linearly dependent if there exists a non-zero scalar k that satisfies vi+kvj = 0 ∀ i 6= j

53


For example, let v1 =(1 2 3

)and v2 =

(2 4 6

). What value of k will satisfy the condition

vi + kvj = 0 ∀ i 6= j?

For example, if the coefficient matrix of a system of equations is

A =

3 4 50 1 26 8 10

=

v1

v2

v3

We find that since (6 8 10) = 2(3 4 5), we have v3 = 2v1 = 2v1 + 0v2. Thus, the third row

is expressible as a linear combination of the first two rows, and the rows are not linearly independent.Alternatively, we may write the previous equation as 2v1 + 0v2 − v3 = 0. Therefore, we find the scalarsk = 2, 0,−1 such that kvi + kvj + kvl = 0, respectively. Therefore, the three rows are not linearlyindependent.

Theorem 4.2. A determinant would exist if and only if all rows and columns of the coefficient matrixare linearly independent

Again, the third row in the previous matrix is only a multiple of the first row. Such linear dependencewill reduce, in effect, the three rows into only two rows with three endogenous variables, which willyield an infinite number of solutions. Therefore, linear independence is a sufficient condition for adeterminant and, thus, for a unique solution to exist. Given a small matrix, row or column independenceis easily verified by visual inspection. However, for a matrix of larger dimension, visual inspection maynot be feasible. Fortunately, we know that a determinant would exist if and only if all rows and columnsof the coefficient matrix are linearly independent. Thus, to check for linear independence, we can simplysolve for the determinant and see if it exists.

4.3.3 Solving for Determinants

The determinant of a matrix A is denoted by |A| or det(A).

For a 2× 2 case, the determinant is the product of the principal diagonal elements less the productof the two off-diagonal elements. Given a general 2× 2 matrix, the determinant is given as |A|

A =

(a11 a12

a21 a22

)|A| = a11a22 − a12a21 (35)

For example, find the determinant of B, |B|

B =

(0 12 3

)|B| = 0 · 3− 1 · 2 = −2

For a 3 × 3 Case, one special case we have is the Basket-weaving Method. The basket-weavingmethod of solving for determinants applies only to 3× 3 matrices. It is the same process as getting thedeterminant of a 2× 2 matrix, except we need to recopy the first and second columns of the matrix tothe right of the original matrix, before doing the basket-weave method.

For example 1 −3 42 5 −1−2 3 −4

→ 1 −3 4

2 5 −1−2 3 −4

· 1 −3

2 5−2 3

54


To do the basket weave method, we first add the diagonals going left to right

(1)(5)(−4) + (−3)(−1)(−2) + (4)(2)(3) = −2

Now we add the diagonal products going right to left

(−2)(5)(4) + (3)(−1)(1) + (−4)(2)(−3) = −19

Take the difference of the two sums to get the determinant

−2− (−19) = 17

For matrices with a higher order, we can use the Laplace Expansion. The Laplace expansion isthe basic technique for calculating determinants and applies to large matrices. To understand this, letus use an example of a general 3× 3 matrix

A =

a11 a12 a13

a21 a22 a23

a31 a32 a33

1. Select any row or column in the matrix (say row 1, but preferably a row or column that has many

0s or 1s to simplify the process).

2. Calculate the minor for each element of the selected row/column

• Since row 1 was selected, we can compute for 3 minors for each element in row 1, i.e.a11, a12, a13.

• The minor of a11, denoted as |M11|, is the determinant of the sub-matrix associated witha11.

• The sub-matrix associated with a11 is composed of the elements of the original matrix ex-cluding the elements in the row and column which a11 appears. The submatrix associatedwith a11 is the following.

a11 sub-matrix =

(a22 a23

a32 a33

)taken from A =

a11 a12 a13

a21 a22 a23

a31 a32 a33

• Thus, the minor of a11 is |M11| = a22a33 − a23a32.

• Do the same to get the minors of a12 and a13 where |M12| = a21a33 − a23a31 and |M13| =a21a33 − a23a31

3. Once you get the minors, you can now calculate the cofactor of each element of the selected rowor column.

• The cofactor of aij is its minor with an assigned algebraic sign.

• The general formula for the cofactor is |Cij | = (−1)i+j |Mij |.• In the forgoing illustration, the cofactors are:

a11 : |C11| = (−1)1+1|M11| = |M11|

a12 : |C12| = (−1)1+2|M12| = −|M12|

a13 : |C13| = (−1)1+3|M13| = |M12|

4. Calculate the determinant using the following formula

|A| =∑

aij · |Cij | = a11 · |C11|+ a12 · |C12|+ a13 · |C13|

55


For example, say you were given with Matrix A

A =

1 0 42 3 53 4 6

We select the row or column with the most zeros. In this case, let us select the first row.

|M11| = (3)(6)− (5)(4) = −2

|M12| = (2)(6)− (5)(3) = −3

|M13| = (2)(4)− (3)(3) = −1

We then calculate for the cofactors

|C11| = (−1)1+1(−2) = −2

|C12| = (−1)1+2(−3) = 3

|C13| = (−1)1+3(−1) = −1

Lastly, we apply the formula for the determinant

|A| = (1)(−2) + (0)(3) + (4)(−1) = −6

Find the determinants of the following matrices

1.

1 −6 52 2 5−1 −4 1

2.

15 4 8−12 −7 5

0 −5 15

3.

2 1 0 41 2 1 40 3 2 22 1 3 3

4.3.4 Basic Properties of Determinants

There are a couple of properties and aspects of determinants we must take note of

1. The interchange of rows and columns does not affect the value of a determinant, i.e. |A| = |AT |

2. The interchange of any 2 rows or any 2 columns will alter the sign but not the numerical value ofthe determinant

3. The multiplication of any 1 row or 1 column by a scalar k will change the value of the determinantk-fold

4. The addition (subtraction) of a multiple of any row (or column) to (from) another row (or column)will leave the value of the determinant unaltered.

5. If at least 1 row (or column) is a multiple of another row (or column), the value of the determinantwill be zero.

56


6. A square matrix is singular if the determinant does not exist (i.e. is equal to zero). In otherwords, a square matrix is singular if one or more of its rows and columns are linearly dependent.On the contrary, a square matrix is non-singular if its determinant exists (i.e. is non-zero).

7. The determinant of a triangular matrix is simply the product of its diagonal entries.

4.3.5 Defining the Inverse of a Matrix

Theorem 4.3. The inverse of a square matrix A is a square matrix B of the same order such thatAB = BA = I where I is an identity matrix

For example, consider these two matrices

A =

(1 −12 0

); B =

(0 1/2−1 1/2

)If you multiply A ·B (

1 −12 0

)·(

0 1/2−1 1/2

)=

(1 00 1

)Similarly, if we do B ·A (

0 1/2−1 1/2

)·(

1 −12 0

)=

(1 00 1

)Therefore, we can consider B as the inverse of A. From the definition, it is clear that if B is an

inverse of A, then A is an inverse of B. However, there are such matrices that do not have inverses.

The inverse of a square matrix is unique. For if B and C are both inverses of A, then it must be thecase that:

AB = I = BA, AC = I = CA

Then, it must also hold that

B = BI = B(AC) = (BA)C = IC = C

Since the inverse of a square matrix is unique, we identify it by denoting it as A−1. Thus,

AA−1 = I = A−1A

4.3.6 Solving for the Inverse of a Matrix

To get the inverse of a matrix, you must first compute for the determinant of the matrix. This impliesthat the matrix must be both square and non-singular.

Let A =

a11 . . . a1m...

. . ....

am1 . . . amm

1. Derive the cofactor of every entry of A

2. Set up the matrix of cofactors

C =

|C11| . . . |C1m|...

. . ....

|Cm1| . . . |Cmm|

3. Take the transpose of matrix C, which is also known as the adjoint of A, i.e. C ′ = adj(A)

57


4. Divide C ′ by the determinant of A (i.e divide the transpose of the matrix of cofactors by thedeterminant of A to get the inverse.

A−1 =C ′

|A|=

|C11| . . . |C1m|...

. . ....

|Cm1| . . . |Cmm|

|A|

=

|C11||A . . . |C1m|

|A|...

. . ....

|Cm1||A| . . . |Cmm|

|A|

As an example, find the inverse of the following matrix.1 0 4

2 3 53 4 6

We first derive the cofactor of every entry in A and set up the cofactor matrix. To get the cofactors,

we have to first get the minors of every entry. The minors when computed are |M11| = −2, |M12| =−3, |M13| = −1, |M21| = −16, |M22| = −6, |M23| = 4, |M31| = −12, |M32| = −3, |M33| = 3.

Therefore, the cofactors are just |C11| = −2, |C12| = 3, |C13| = −1, |C21| = 16, |C22| = −6, |C23| =−4, |C31| = −12, |C32| = 3, |C33| = 3. This allows us to build the cofactor matrix.

C =

−2 3 −116 −6 −4−12 3 3

To obtain the adjoint, we simply take the transpose of the cofactor matrix, that is, adj(A) = C ′

adj(A) = C ′ =

−2 16 −123 −6 3−1 −4 3

Finally, because the determinant of matrix A is -6 (which you can compute yourself), the inverse of

matrix A is

A−1 =adj(A)

|A|=

−2 16 −123 −6 3−1 −4 3

−6

=

1/3 −8/3 2−1/2 1 −1/21/6 2/3 −1/2

Find the inverses of the following matrices

1.(

3 21 0

)

2.

4 1 −10 3 33 0 7

3.

1 −1 21 0 34 0 2

Note that inverse matrices have a couple of properties

• (A−1)−1 = A (The inverse of an inverse is the original)

58


• (AB)−1 = B−1A−1

• (AT )−1 = (A−1)T

Moreover, a matrix has to have a non-zero determinant in order for us its inverse to exist. Therefore,a matrix is nonsingular if and only if its determinant is non-zero (i.e. if it has an inverse). On the otherhand, a matrix is singular if and only if its determinant is zero (i.e. if it does not have an inverse).

4.3.7 Solving a System of Equations by Matrix Inversion

You can use matrix inversion to find the solution vector to a system of linear equations. Recall thatour equation system can be expressed in the form Ax = d, where A is the matrix of coefficients, x isthe column vector of endogenous variables, and d is the column vector of constant terms or exogenousvariables. To solve for the column vector x, which will give us the solution values of the endogenousvariables, we can pre-multiply A−1 to both sides of Ax = d to solve for x. That is,

A−1Ax = A−1d

x = A−1d (36)

For example, consider the system of linear equations we have below3x1 + x3 = 4

2x1 + x2 + 4x3 = 6

x1 + x3 = 8

Rewritting in Matrix Form 3 0 12 1 41 0 1

x1

x2

x3

=

468

To isolate x, we pre-multiply both sides of Ax = d by A−1 and get x = A−1d. We solve first for

A−1 (the inverse of A). Afterwards, we getx1

x2

x3

︸︷︷︸

x

=

0.5 0 −0.51 1 −5−0.5 0 1.5

︸︷︷︸

A−1

468

︸︷︷︸d

=

(0.5)(4) + (0)(6) + (−0.5)(8)(1)(4) + (1)(6) + (−5)(8)

(−0.5)(4) + (0)(6) + (1.5)(8)

=

−2−3010

Therefore, x1 = −2, x2 = −30, x3 = 10.

Write the following systems of equations in matrix form, then find the solution values usingmatrix inversion.

1.

Qd = 500− 50P

Qs = 50 + 25P

Qd = Qs

2.

x− 2y + 3z = 7

2x+ y + z = 4

−3x+ 2y − 2z = −10

3.

C = 260 + 0.4Y

I = 300 + 0.5Y

Y = C + I + 110

59


4.3.8 Solving a System of Equations using Cramer’s Rule

Cramer’s Rule is an alternative approach for solving systems of equations. Again, a non-zero deter-minant is necessary when using Cramer’s Rule; hence, the matrix must be both square and nonsingular.

The procedure for Cramer’s Rule is as follows

1. Say we are solving for x1 first. Given Ax = d, substitute the d column vector into the first columnof the A matrix. Let the resulting matrix be A1.(

a11 a12

a21 a22

)(x1

x2

)=

(d1

d2

)→(d1 a12

d2 a22

)2. Derive the determinant of A1 and divide it by the determinant of A. This will yield the solution

value for x1.

A1 =

(d1 a12

d2 a22

)→ |A1| = d1a22 − d2a12

A =

(a11 a12

a21 a22

)→ |A| = a11a22 − a21a12

x1 =|A1||A|

=d1a22 − d2a12

a11a22 − a21a12

3. Do the same to get x2 (substitute first the d column vector into the second column of the Amatrix).

Consider again our example from matrix inversion3x1 + x3 = 4

2x1 + x2 + 4x3 = 6

x1 + x3 = 8

Rewriting this in matrix form, we have3 0 12 1 41 0 1

x1

x2

x3

=

468

Solving first for x1, we transfer the d column vector into the first column of the A matrix, and name

the resulting matrix A1 3 0 12 1 41 0 1

→ A1 =

4 0 16 1 48 0 1

We get the determinant of A1 which is −4. Dividing the determinant of A1 by the determinant of

A yields the solution value for x1. The determinant of A is 2.

x1 =|A1||A|

=−4

2= −2

Solving first for x2, we transfer the d column vector into the second column of the A matrix, andname the resulting matrix A2 3 0 1

2 1 41 0 1

→ A2 =

3 4 12 6 41 8 1

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We get the determinant of A2 which is −60. Dividing the determinant of A2 by the determinant ofA yields the solution value for x2. The determinant of A is 2.

x2 =|A2||A|

=−60

2= −30

Solving first for x3, we transfer the d column vector into the second column of the A matrix, andname the resulting matrix A3 3 0 1

2 1 41 0 1

→ A3 =

3 0 42 1 61 0 8

We get the determinant of A3 which is 20. Dividing the determinant of A3 by the determinant of A

yields the solution value for x3. The determinant of A is 2.

x3 =|A3||A|

=20

2= 10

Notice that the same answer should be obtained whether you are using matrix inversion or Cramer’srule.

Write the following systems of equations in matrix form, then find the solution values usingCramer’s Rule.

1.

Qd = 500− 50P

Qs = 50 + 25P

Qd = Qs

2.

x− 2y + 3z = 7

2x+ y + z = 4

−3x+ 2y − 2z = −10

3.

C = 260 + 0.4Y

I = 300 + 0.5Y

Y = C + I + 110

5 Economic Applications of Matrix Algebra and Comparative Statics

5.1 Single-Commodity Competitive Market

Partial equilibrium models, like those discussed previously, can be transformed into matrices and solvedwith ease by matrix inversion or Cramer’s Rule.

For example, consider a competitive market for a single commodity. A market is competitive if allfirms producing the commodity are price-takers, i.e. the firms take the market price as given and selltheir products at that price. Hence,

Qd = a− bP (a > 0, b > 0)

Qs = −c+ dP (c > 0, d > 0)

In the forms above, Qd is the quantity demanded, Qs is the quantity supplied, P is price, and a, b, c,and d are parameters. The restrictions on the parameters b and d indicate the usual laws of demandand supply; if price goes down, buyers will try to buy more and if price goes up, sellers are likely tosell more. The price at which buyers want to buy exactly as much as sellers want to sell is called the

61


equilibrium price, i.e. the price where Qd = Qs.

In this case, we say that the market clears and refer to the equilibrium condition Qd = Qs as amarket-clearing condition. The common value Q∗ = Qd = Qs is called the equilibrium quantityand (P∗, Q∗) an equilibrium point. Mathematically and graphically, the intersection of the demand andsupply functions correspond to the equilibrium point and satisfies the market clearing conditions.

Demand Function: Qd = a− bP (a > 0, b > 0)

Supply Function: Qs = −c+ dP (c > 0, d > 0)

Market Clearning Condition: Qd = Qs

5.2 National Income Model

Vectors and matrices are also useful in models of the national economy of the linear variety. To illustrate,we consider a simple national income model given by the equations:

C = C0 + b(Y − T ) (C0 > 0, 0 < b < 1)

T = T0 + tY (T0 > 0, 0 < t < 1)

Y = C + I0 +G0 (I0 > 0, G0 > 0)

In the model above, Y denotes national income, C denotes aggregate consumption expenditure, Tis tax revenue, I0 is investment, and G0 is government expenditure. In this model, C, T, and Y areendogenous variables, I0 and G0 are exogenous variables, and C0, T0, b, and t are parameters.

The first equation in the system is a consumption function, where b is the marginal propensity toconsume out of disposable income Y −T . The restriction that 0 < b < 1 simply says that some amountbut not all of disposable income is spent on consumption. It also assumes that if disposable income iszero, autonomous consumption C0 is positive, made possible by borrowing.

The second equation in the system is a tax revenue equation, where the income tax rate t is apositive fraction less than 1 to indicate that some, but not all, of income is collected as income tax. T0

is assumed positive since there are other sources of tax revenues other than income.

The third equation in the system is the equilibrium condition which states that national income Yis equal to national total expenditure (C + I0 +G0).

Assume that GDP Y is equal to 1, 200. Consumption C is given by the equation C = 125 +0.75(Y − T ). Investment I is given by the equation I = 200 − 10r, where r is the real interestrate. Taxes T are 100 and government spending G is 150. What are the equilibrium values ofC, I, and r?

5.3 Closed Economy IS-LM Model

Consider a model of an economy consisting of a goods market and a money market. In the goodsmarket, consumption C is a linear function of national income Y

C = C0 + b(1− t)Y (0 < b, t < 1)

In the equation above, t is income tax rate and b is the marginal propensity to consume out ofdisposable income (1− t)Y . Investment is a linear function of interest rate R

I = I0 − cR (0 < c < 1)

Equilibrium in the goods market requires

62


Y = C + I +G0

In the aforementioned system of three equations, Y,C, I, and R are endogenous variables, G0 is theexogenous variable, and C0, b, t, I0, and c are parameters.

In the money market, the demand for money Md is given by

Md = kY − lR (k, l > 0)

We assume that the quantity of money supplied Ms is determined by the central bank and is,therefore, exogenous and equal to M0 . The equilibrium condition in the money market is

Md = Ms = M0

In the form above, M0 is exogenous and k and l are parameters.Moreover, this implies that

kY − lR = M0

In IS-LM analysis, we solve for the income Y and the interest rate R that will clear both marketssimultaneously, i.e. both markets are simultaneously in equilibrium. Solving the system of equationsabove yield us the IS and LM equation below.

Y =C0 + I0 − CR+G0

1− b(1− t)

Y =M0 + lR

k

When we equate the IS equation which shows equilibrium in the goods market, with the LM equationwhich shows equilibrium in the money market, we can solve for the interest rate R and, eventually, theGDP level Y where both the goods and money markets are in equilibrium.

Let the IS equation be

Y =A

1− b− g

1− bR

Let the LM equation be

Y =M0

k+l

kR

If b = 0.7, g = 100, A = 252, k = 0.25, l = 200, and M0 = 176

• Write the IS-LM system in matrix form

• Solve for Y and R using matrix inversion and Cramer’s Rule.

5.4 Leontief Input-Output Model

Input-output analysis is concerned with the relationships among the industries of an economy. Theeconomy consists of industries whose outputs are produced using inputs from the industries (calledintermediate inputs) and inputs from sectors outside the industries (called primary inputs). Forexample, a farmer produces rice by using rice seeds, fertilizer, and fuel obtained from other indus-tries(intermediate inputs) and land and labor (primary inputs).

The output of an industry is, therefore, demanded as inputs to the production of the other in-dustries and itself. When outputs are demanded for the production of more outputs, these demands

63


by the industries are called intermediate demands. Outputs that are demanded for final consump-tion by sectors outside the industries such as households and the government are called final demands.

For example, consider an economy with 3 production sectors: Agriculture, Industry, and Services.The table to follow shows a 3-sector input-output table of the Philippine economy in 19941. The rowsrepresent the producing sector of the economy, and the columns represent the consuming sector of theeconomy. The Total Output column has the same entries as the Total Input row since thevalue of total output equals the value of total inputs used to produced that output. All data are inbillions of pesos.

Intermediate Demand Final Demand Total OutputAgri Industry Services

Intermediate InputAgri 44.7 214.5 15.8 197.3 472.3Industry 42.8 210.1 174.2 861.9 1289Services 28.6 149.9 290.8 1105 1574.3

Primary Input 356.2 714.5 1093.5Total Input 472.3 1289 1574.3

Table 1: 1994 Philippine Input Output Table

In the column market Agri (for Agriculture), it says that the 472.3 billion pesos worth of output ofAgriculture was produced by using 44.7 billion pesos worth of products from Agriculture, 42.8 billionpesos worth of products from Industry, 28.6 billion pesos worth of products from Services, and 356.2billion pesos worth of primary inputs.The other two columns (Industry and Services) are interpreted ina similar way.

The row for Agriculture says that the total Agriculture output of 472.3 billion pesos, 33.7 billionwas purchased by Agriculture itself to produce its output, 214.5 billion pesos was purchased by Servicesto produce its output.,15.8 billion pesos was purchased by Services to produce its output, and 197.3billion pesos was purchased by the Final Demand Sector. The other two rows are interpreted similarly.

From the previous table, we create a new table (see below) by dividing by total input every entry inits column. The results are called technical coefficients. The first column says that producing 1 pesoworth of output in Agriculture requires 9 centavos worth of agricultural products, 9 centavos worth ofIndustrial products, 6 centavos worth of services, and 75 centavos worth of primary inputs. The 3× 3matrix of coefficients above the Primary Input row is usually called the technology matrix and isdenoted by A.

Agri Industry ServicesAgri 0.095 0.167 0.010Industry 0.091 0.163 0.111Services 0.060 0.116 0.185Primary Input 0.754 0.554 0.694Total Input 1.000 1.000 1.000

Table 2: 1994 Technology Matrix

5.4.1 The Leontief Equation

To formulate the Leontief model, we use the following notations:

• xi output level of sector i

• xij output level of sector i supplied to sector j1Taken from Danao (2011) Core Concepts of Calculus

64


• aij output level of sector i supplied to sector j to produce 1 unit of output j.

• di final demand on sector i

• zkj amount of primary input k in sector j

• bkj amount of primary input k needed to produce one unit of output j

Thus, with these notations, our two tables (the input-output table and the technology coefficientmatrix) would, respectively, look like

Intermediate Demand Final Demand Total OutputAgri Industry Services

Intermediate InputAgri x11 x12 x13 d1 x1

Industry x21 x22 x23 d2 x2

Services x31 x32 x33 d3 x3

Primary Input z11 z12 z13

Total Input x1 x2 x3

Table 3: Mathematically Representing the IO Table

Agri Industry ServicesAgri a11 a12 a13

Industry a21 a22 a23

Services a31 a32 a33

Primary Input b11 b12 b13

Total Input x1 x2 x3

Table 4: Mathematically Representing the Technology Matrix

The technical coefficients aij are defined as

aij =xijxj

hence, xij = aijxj

If each sector produces an output just enough to meet intermediate and final demands, then, fromour input-output table above we get the following equations

x1 = a11x1 + a12x2 + a13x3 + d1

x2 = a21x1 + a22x2 + a23x3 + d2

x3 = a31x1 + a32x2 + a33x3 + d3

In matrix form, we have x1

x2

x3

=

a11 a12 a13

a21 a22 a23

a31 a32 a33

x1

x2

x3

+

d1

d2

d3

Hence

x = Ax+ d→ x−Ax = d→ (I −A)x = d

Therefore, thee Leontief equation is stated as

(I −A)x = d (37)

In the equation above, we denote (I − A) as the Leontief matrix. For a given vector d of finaldemands and a given technology matrix A, the vector of outputs x may be obtained from the modifiedLeontief equation

65


x = (I −A)−1d

Given the Leontief equation x = (I−A)−1d , which yields the vector of final outputs produced by allindustries, the matrix (I −A)−1 is called the Leontief inverse matrix. To interpret the entries of thematrix, suppose that the final demand for product 1 is increased by one unit while the final demandsof the other products remain unchanged. Let the change in the final demand vector be denoted by ∆d,i. e.

∆d =

100

Corresponding to this change in final demand, there is a change in final output ∆x. Hence,

x+ ∆x = (I −A)−1(d+ ∆d)

x+ ∆x = (I −A)−1d+ ∆x = (I −A)−1∆d

Subtracting the Leontief Inverse equation x = (I −A)−1d from the equation above, we get

∆x = (I −A)−1∆d

For convenience, let L = (I −A)−1. Therefore, ∆x = L(∆d)∆x1

∆x2

∆x3

=

L11 L12 L13

L21 L22 L23

L31 L32 L33

100

=

L11

L21

L31

Thus, the first column of (I −A)−1 represents the increases in the outputs of the various industries

due to a unit increase in the final demand of industry 1, holding the final demands in the other industriesconstant. The second and third columns of the Leontief inverse matrix are interpreted in a similar way.

5.4.2 Demand for Primary Inputs

For convenience, assume that there are only 2 primary inputs and define. The amount of primary inputk used to produce 1 unit of product j is Bkj . The total amount of primary input k needed for productionis yk. The coefficients bkj are the primary input coefficients. Note that bkjxj is the amount of primaryinput k needed to produce output xj . Thus, the total amount of primary input 1 needed for productionis

y1 = b11x1 + b12x2 + b13x3

and the total amount of primary input 2 needed for production is

y2 = b21x1 + b22x2 + b23x3

Combining these two equations in matrix form, we have

y =

(b11 b12 b13

b21 b22 b23

)︸︷︷︸

B

x1

x2

x3

︸︷︷︸

x

The equation y = Bx gives the total amount of primary inputs needed to produce the outputs ofall sectors.

66


Consider an economy divided into an agricultural sector (A) and an industrial sector (I). Toproduce one unit in sector A requires 1/6 unit from A and 1/4 unit from I. To produce one unitin sector I requires 1/4 unit from A and 1/4 unit from I. Suppose final demands in each of thetwo sectors are 60 units.

• Write down the input-output matrix equation for this economy.

• Find the number of units that have to be produced in each sector in order to meet the finaldemands.

Consider the following technology matrix A, primary input coefficients matrix B, and the finaldemand vector d

A =

0.16 0.17 0.010.12 0.41 0.180.04 0.07 0.17

, d =

107437538

, B =(0.68 0.35 0.64

)• What is the vector of outputs produced?

• If the final demand of industry 2 is increased by 1 unit while keeping the final demands ofthe other industries fixed, by how much will the outputs of all industries change?

• Suppose that the estimated available supply of primary inputs is 1200. Can this meet therequired outputs?

5.5 Comparative Statics

Economic analysis is often concerned with determining the effects of certain public policies on economicvariables (e.g. equilibrium price and quantity, GDP, money demand). The effects are usually obtainedby comparing the equilibrium values without the policy with those resulting from the given policy.

In mathematical terms, the solution, representing a decision variable (e.g. equilibrium price), ofan equation or a set of equations depends on a parameter or an exogenous variable. A change in theparameter or exogenous variable can represent a certain policy. The technique for determining theresponse of a decision variable to a change in parameter or exogenous variable is called comparativestatics.

5.5.1 The Jacobian Matrix

The Jacobian Matrix is a matrix of all first-order partial derivatives of a system of equations. Sincefirst-order derivatives (or partial derivatives) inform us what the change in the decision variable is dueto a policy change (i.e. change in the exogenous variable), then the Jacobian Matrix is useful for com-parative static analyses.

Consider the two functions

y1 = 2x1 + 3x2

y2 = 4x21 + 12x1x2 + 9x2

2

If we get all four partial derivatives

∂y1

∂x1= 2,

∂y1

∂x2= 3,

∂y2

∂x1= 8x1 + 12x2,

∂y2

∂x2= 12x1 + 18x2

We can arranged these partial derivatives into a square matrix in a prescribed order. Doing so yieldsus the Jacobian Matrix, denoted by J , for short.

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J =

(∂y1∂x1

∂y1∂x2

∂y2∂x1

∂y2∂x2

)=

(2 3

8x1 + 12x2 12x1 + 18x2

)It is also possible to get the determinant of the Jacobian matrix, denoted by |J |, for short. Given a

system of implicit functions:

F 1(y1, . . . , yn, x1, . . . , xk) = 0

...

Fn(y1, . . . , yn, x1, . . . , xk) = 0

Where y1, . . . , yn are endogenous variables and x1, . . . , xk are exogenous variables. Using thesesystem of implicit functions, a Jacobian matrix can be formed. To derive the Jacobian matrix from asystem of implicit functions, the following steps must be followed.

1. Take the total differential of all implicit functions in the system.

2. Move all partial derivatives associated with the exogenous variables to the right-hand side of theequations.

3. Arrange the results in matrix format.F1y1 . . . F 1

yn...

. . ....

Fny1 . . . Fnyn

︸︷︷︸

Jacobian Matrix

·

dy1...dyn

︸︷︷︸dY matrix

=

−F1x1dx1 − F 1

x2dx2 − . . .− F 1xkdxk

...−Fnx1dx1 − Fnx2dx2 − . . .− Fnxkdxk

︸︷︷︸

dx matrix

Say, due to a policy change, there is a change in only x1, i.e. dx2 = . . . = dxk = 0. Therefore,−F1x1dx1 − F 1

x2dx2 − . . .− F 1xkdxk

...−Fnx1dx1 − Fnx2dx2 − . . .− Fnxkdxk

collapses to

−F1x1dx1...

−Fnx1dx1

This leaves us with the followingF

1y1 . . . F 1

yn...

. . ....

Fny1 . . . Fnyn

·dy1

...dyn

=

−F1x1dx1...

−Fnx1dx1

Divide both sizes by dx1, a scalar, we get

F1y1 . . . F 1

yn...

. . ....

Fny1 . . . Fnyn

·

∂y1∂x1...

∂yn∂xn

=

−F1x1...

−Fnx1

You can now derive the ∂yi

∂xi∀ i = 1, 2, ..., n using matrix inversion or Cramer’s rule.

For example, say you were given with the following system of implicit functionsxy − w = 0

y − w3 − 3z = 0

w3 + z3 − 2zw = 0

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The endogenous variables are x, y, and w are endogenous variables and z is an exogenous variable.The solution values for the system of equations above is x = 0.25, y = 4, w = 1, and z = 1. Supposeyou wanted to determine the changes in the endogenous variables when z changes.

To do this, we first take the total differential of the system.

ydx+ xdy − dw = 0

dy − 3w2dw − 3dz = 0

(3w2 − 2z)dw + (3z2 − 2w)dz = 0

Then we move all partial derivatives associated with z to the right-hand side.

xdx+ ydx− dw = 0

dy − 3w2dw = 3dz

(3w2 − 2z)dw = −(3z2 − 2w)dz

We then arrange the system in matrix formx y −11 0 −3w2

0 0 (3w2 − 2z)

dydxdw

=

03dz

(2w − 3z2)dz

Substituting the solution values into the Jacobian matrix and the dx matrix, and dividing both sizes

by dz, we get 14 4 −11 0 −30 0 1

dydzdxdzdwdz

=

03−1

We can now solve for dy

dz ,dxdz and dw

dz using Cramer’s Rule.

dy

dz=

det

0 4 −13 0 −3−1 0 1

det(J)

=−24

−4= 6

dy

dz=

det

0.25 0 −11 3 −30 −1 1

det(J)

=1

−4= −1

4

dy

dz=

det

0.25 4 01 0 30 0 −1

det(J)

=4

−4= −1

Consider the following national income model

C = C0 + b(Y − T )

T = T0 + tY

Y = C + I0 +G0

What would be the effect of a fiscal policy (i.e. a change in government spending G0 on theequilibrium values of C, T, and Y .

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6 Multivariate Unconstrained Optimization

We discussed a lot of the concepts of unconstrained optimization in the section on differentiation. Thissection will deal on specific cases of optimization which involve more than one variable.

6.1 Topological Underpinnings of the First Derivative

Economic problems are frequently concerned with making choices from among several alternatives onthe basis of an objective. For example, a firm chooses the amounts of inputs and output that minimizecosts or a consumer chooses the amount of each good to purchase that will maximize utility. Suchproblems may formally be analyzed by generalizing the ideas and techniques employed for functions ofa single variable.

The definitions of a global optima and local optima of functions of several variables are the same asthose for functions of a single variable.

• If f is a function whose domain D is a subset of Rn, then a point x∗ in D is a global maximizerof f if and only if f(x) ≤ f(x∗) for all x in D.

• f f is a function whose domain D is a subset of Rn, then a point x∗ in D is a global minimizerof f if and only if f(x) ≥ f(x∗) for all x in D.

• If f is a function whose domain D is a subset of Rn, then a point x∗ in D is a local maximizerof f if and only if there is an open neighbourhood N centred at x∗ such that f(x) ≤ f(x∗) for allx in N .

• If f is a function whose domain D is a subset of Rn, then a point x∗ in D is a local minimizerof f if and only if there is an open neighbourhood N centred at x∗ such that f(x) ≥ f(x∗) for allx in N .

The concept for the First-Derivative Test is vital in our understanding of identifying extrema.Consider the function f(x, y) below. Suppose that f attains a local maximum at (x∗, y∗) so that P isthe highest point on the surface over an open disk centred at (x∗, y∗). Any movement from (x∗, y∗) inany direction will decrease the value of f. In particular, this is true for movements along AB (parallelto the x-axis), i.e. P is the highest point on the curve APB. This means that the slope of APB at(x∗, y∗) is zero, i.e. . fx(x∗, y∗) = 0.

Figure 28: Topologically Showing a Local Extremum

Similarly, along the line CD (parallel to the y-axis), the slope of the curve CPD at (x∗, y∗) is alsozero, i.e. fy(x∗, y∗) = 0.

If (x∗, y∗) is an interior point of the domain of f and (x∗, y∗) is a local minimizer, then by the sameargument as above, the first partial derivatives of f at (x∗, y∗) are equal to zero. Thus, at local optimiz-ers, it is necessary that the partial derivatives are equal to zero. This result can be generalized tofunctions of more than two variables.

70


Theorem 6.1. A point (x, y) is called a stationary point of f if and only if fx(x, y) = fy(x, y) = 0.

Again, the first-derivative test above is only a necessary but not sufficient condition for iden-tifying local maxima or minima. For example, say you were given with the function y = f(x1, x2) =2x2

1−x1x2 + 2x22 + 10. Find the local optima of the function using the First Derivative Test. We obtain

the first order partial derivatives

fx1 = 4x1 − x2, fx2 = −x1 + 4x2

The first-order condition for a local optima requires fx1 = fx2 = 0.

0 = 4x1 − x2, 0 = −x1 + 4x2

Solving both equations simultaneously yields x1 = 0 and x2 = 0 . But we still don’t know whetherboth are local maxima or minima.

Identify the local optima of f(x, y) = x2 + y2 + xy + x+ 5y using the First Derivative Test.

Assume that the output price p, wage rate w, and rate of return to capital r are all given andequal to 1. Suppose further that output q is produced using inputs of labor l and capital k. Theproduction function is represented by q = l0.25k0.25. Identify the local optima that potentiallymaximizes profit.

A word of caution! There are stationary points that are not optimizers. For example, consider thefunction

f(x, y) = x2 − y2

The first-order partial derivatives are

fx = 2x, fy = −2y

Thus, (x∗, y∗) = (0, 0) is a stationary point of f . But on the x-axis where y = 0, f(x, y) = x2; hence,any movement away from the origin along the x-axis will increase f(x, y). On the other hand, alongthe y-axis where x = 0, f(x, y) = −y2; hence, any movement away from the origin along the y-axis willdecrease f(x, y). Thus, (x∗, y∗) = (0, 0) is neither a local maximizer nor a local minimizer. This point iscalled a saddle point. The name comes from the fact that the shape of the graph looks like a saddle.

Figure 29: Illustrating a Saddle Point Equilibrium

71


6.2 Second Derivative Test

The first-order conditions merely identify stationary points and these can be minimizers, maximizers,or saddle points. Similar to functions of one variable, there are some sufficient conditions using second-order partial derivatives that could help classify the stationary points.

Consider again the function f(x, y) depicted before where (x∗, y∗) is a local maximizer. The first-order necessary conditions for optimality are

fx(x∗, y∗) = 0, fy(x∗, y∗) = 0

On the curve APB, the second-order condition for a local maximum holds at fxx(x∗, y∗) < 0. Onthe curve CPD, the second-order condition for a local maximum holds at fyy(x∗, y∗) < 0.

Thus, on the curves defined by lines parallel to the axes, these two conditions guarantee that (x∗, y∗)is a local maximizer on these two curves. However, these two conditions have nothing to say about thebehaviour of a curve on the surface defined by lines going through (x∗, y∗) not parallel to any of theaxes. In other words, these two conditions do not fully capture the behaviour of the function at otherpoints in the neighbourhood of (x∗, y∗), i.e. they are not sufficient to identify a local maximizer.Whatwe need are additional conditions to ensure that (x∗, y∗) is a maximizer in its neighbourhood. Notsurprisingly, these additional conditions involve the cross-partial derivatives.

6.2.1 Hessian Matrix

The Hessian Matrix is a square matrix of second-order partial derivatives. It describes the localcurvature of a function with more than one variable, and is, therefore, useful for identifying whetherlocal optima are maximizers or minimizers. For a function with two variables x and y, the Hessianmatrix looks like

H(x, y) =

(∂2f∂x2

∂2f∂x∂y

∂2f∂y∂x

∂2f∂y2

)=

(fxx(x, y) fxy(x, y)fyx(x, y) fyy(x, y)

)For a function with n variables, the 2× 2 Hessian matrix can be generalized to look like

H(x1, ..., xn) =

∂2f∂x21

∂2f∂x1∂x2

. . . ∂2f∂x1∂xn

∂2f∂x2∂x1

∂2f∂x22

. . . ∂2f∂x2∂xn

......

. . ....

∂2f∂xn∂x1

∂2f∂xn∂x2

. . . ∂2f∂x2n

6.2.2 Definiteness and Principal Minors

Let A be an n×n matrix. The submatrix obtained by deleting any (n−r) rows and the same numberedcolumns of A is called a principal submatrix of order r, and its determinant is called a principalminor of order r. The principal submatrix of order r obtained by deleting the last (n − r) rows and(n− r) columns of A is called a leading principal submatrix of order r, and its determinant is calleda leading principal minor of order r.

For example, consider

A =

3 0 41 1 21 −2 2

72


The principal submatrices of order 2 are:(1 2−2 2

),

(3 41 2

),

(3 01 1

)The first one is obtained by deleting the 1st row and 1st column, the second one is obtained by

deleting the 2nd row and 2nd column, and the third one is obtained by deleting the 3rd row and 3rdcolumn.

The principal submatrices of order 1 are

(3), (1), (2)

The first one came from deleting rows 2 and 3 and columns 2 and 3.The second one came fromdeleting rows 1 and 3 and columns 1 and 3. And the third one is obtained by deleting rows 1 and 2 andcolumns 1 and 2.

The principal submatrix of order 3 is the matrix A itself.

On the other hand, the leading principal submatrices are only

(3),

(3 01 1

),

3 0 41 1 21 −2 2

The determinants of these leading principal submatrices are called the leading principal minors.

We use these principal minors to determine definiteness. To understand this, let A be a symmetricmatrix

Theorem 6.2. A is positive semidefinite if and only if its principal minors are nonnegative.

Theorem 6.3. A is negative semidefinite if and only if its principal minors of order 1 are nonpositive,those of order 2 are nonnegative, those of order 3 are nonpositive, and so on.

Theorem 6.4. A is positive definite if and only if its leading principal minors are positive.

Theorem 6.5. A is negative definite if and only if its leading principal minors of order 1 is negative,those of order 2 are positive, those of order 3 are negative, and so on.

Decide whether the following matrices are positive definite, negative definite, or neither

1.

−3 0 00 −2 00 0 −1

2.(

6 44 5

)

3.

1 2 32 0 23 2 1

4.

2 −1 −1−1 2 −1−1 −1 2

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6.2.3 Determining the Multivariate SOC

Given a function of two variables x and y, one can form a 2× 2 Hessian matrix using the second-orderpartial derivatives as follows

H(x, y) =

(fxx(x, y) fxy(x, y)fyx(x, y) fyy(x, y)

)Note that fxy = fyx according to Young’s Theorem.

1. If H is negative definite (i.e. fxx < 0 and fxxfyy − f2xy > 0), then (x∗, y∗) is a local maximizer of

f .

2. If H is positive definite (i.e. fxx > 0 and fxxfyy − f2xy > 0), then (x∗, y∗) is a local minimizer of f

3. If fxxfyy − f2xy < 0, then (x∗, y∗) is a saddle point of f

Let f be a function of x= [x1, x2, ..., xn] and suppose that f has continuous second-order partialderivatives. Let x∗ be a stationary point of f .

1. If the Hessian Matrix H is negative definite, then x∗ is a local maximizer of f .

2. If the Hessian Matrix H is positive definite, then x∗ is a local minimizer of f .

3. If the Hessian Matrix H is indefinite, then x∗ is a saddle point of f .

Consider our example from before, say you were given with the function y = f(x1, x2) = 2x21 −

x1x2 + 2x22 + 10. Find the local optima of the function using the First Derivative Test and determine if

this is a local maximizer/minimizer/saddle point. We obtain the first order partial derivatives

fx1 = 4x1 − x2, fx2 = −x1 + 4x2

The first-order condition for a local optima requires fx1 = fx2 = 0.

0 = 4x1 − x2, 0 = −x1 + 4x2

Setting the first-order partial derivatives equal to zero, we get the stationary point (x∗, y∗) = (0, 0).The Hessian matrix of f is

H =

(4 −1−1 4

)The principal minors of H of order one are 4 and 4; the principal minor of order two is 15 (i.e.

16− 1 = 15). Therefore, H is positive definite. Hence, (x∗, y∗) = (0, 0) is a local maximizer of f .

Identify the local optima of the following functions using the First Derivative Test and SecondDerivative Test

1. f(x, y) = x2 − y2 + xy

2. f(x, y) = x2 + y2 + xy + x+ 5y

3. f(x, y, z) = x2 + xz − y + y2 + yz + 3z2

4. y = 20x21 + 10x1x2 + 40x2

2 + 10x1x3 + 10x23 + 20

A firm produces two goods in pure competition and has the following total revenue and total costfunctions below. Find the optimal Q1 and Q2 which maximizes profit. Use the Hessian Matrixto determine that profit is indeed maximized.

TR = 15Q1 + 18Q2, TC = 2Q21 + 2Q1Q2 + 3Q2

2

74


6.2.4 Concavity and Convexity

Let f(x) have continuous second-order partial derivatives

• The function f is concave or weakly concave if and only if the Hessian matrix of f is negativesemidefinite.

• The function f is convex or weakly convex if and only if the Hessian matrix of f is positivesemidefinite.

• If the Hessian matrix of f is negative definite for all x in the set, then f is strictly concave.

• If the Hessian matrix of f is positive definite for all x in the set, then f is strictly convex.

Determine the concavity or convexity of the following functions

• f(x, y) = x0.5y0.5

• f(x, y) = x2 − xy + 3y2

6.3 Comparative Statics and Specific Functional Forms

When performing optimization problems, we are often concerned with how the optimal values of vari-ables will change due to changes in the parameters. This is called comparative static analysis.Iffunctions have specific functional forms, we can always find the optimal values of the variables usingthe First and Second Derivative Tests, then perform comparative statics on these variables.

For example, consider the problem of maximizing

z = f(x, y, a) = −x2 − y2 + a(x+ y), where a > 0

where a is a parameter. The first-order conditions are

fx = −2x+ a = 0, fy = −2y + a = 0

Solving both first-order conditions yields the stationary point (x∗, y∗) =(a2 ,

a2

). The Hessian matrix

of the function is

H =

(−2 00 −2

)This matrix is a negative definite matrix, hence, the function is strictly con cave and (x∗, y∗) is a

local maximizer. The maximum z is therefore

z = −x2 − y2 + a(x+ y)

z∗ = −(a/2)2 − (a/2)2 + a[(a/2)(a/2)]

z∗ = −(a2/4)− (a2/4) + a2 = a2/2

Now, by how much will our choice variables z∗, x∗, and y∗ change when a changes marginally? Whatis the direction of the change?

Performing comparative static analysis, we get

∂z∗

∂a= a > 0,

∂x∗

∂a=

1

2> 0,

∂z∗

∂a= a > 0

Thus, as a increases, z∗, x∗, and y∗ also increase.

75


Consider the profit-maximizing firm whose profit function is π(q) = pq − aq2 − bq − tq, wherea, b, t > 0. What is the marginal effect of a change in the tax rate t on the optimal outputproduced?

7 Constrained Optimization

Our previous topic talked on optimization without any constraints. That is, the objective was to max-imize or minimize the objective function y = f(x1, x2) where we have to select x1 and x2 to derive themaximum or minimum value of y. What happens when a constraint is introduced?

Suppose the constraint is g = g(x1, x2).

• The freedom to select your x1 and x2 becomes limited because of the constraint.

• The points that satisfy the constraints are called feasible solutions or feasible points and theset of feasible solutions is called the feasible set or feasible region.

• A feasible solution that maximizes or minimizes the objective function is called an optimalsolution (maximizer or minimizer).

The distinction between unconstrained and constrained optimization is illustrated below. The con-strained optimization problem is

max f(x, y) subject to ax+ by = c

Figure 30: Topological Difference Between Constrained and Unconstrained Maxima

The surface represents the graph of f(x, y). Without any constraint, the maximum occurs at A.With the constraint, the feasible set is restricted to the points on the line L, whose equation is ax+by = c.Thus, maximization is restricted to the curve BCD where the maximum occurs at C.

7.1 Optimization with One Equality Constraint: The Two-Variable Case

The two-variable optimization problem with one equality constraint is given by

max(min)f(x, y) s.t. g(x, y) = b

We discuss two methods of solving this problem. First, the method by substitution. Second, theLagrange multiplier method.

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7.1.1 The Substitution Method

When the constraint g(x, y) = b is simple and it is possible to solve for one of the variable in terms ofthe other, say y = h(x). Then we may substitute this into the objective function. Thus, the problemreduces to optimizing f(x, h(x)) as an unconstrained problem. For example, consider the constrainedoptimization problem below

minx2 + y2 s.t. x+ y = 2

From the constraint, we can solve for y as a function of x

y = h(x) = 2− x

Substitute this into the objective function

f(x) = x2 + (2− x)2

f(x) = 4x− 4

The problem now becomes an unconstrained optimization problem. Using the first derivative test,the stationary point is x∗ = 1.

The second derivative test yields f ′′(x) = 4 > 0. Since this is true for all x, then the function isstrictly convex. Hence, x∗ = 1 is indeed a local minimizer of the function f .

Solving for y∗, we get

f(x) = x2+)2− x)2

f(x) = (1)2 + (2− 1)2

f(x) = 2

Thus, the minimum value of the function is 2.

However, the substitution method has severe dimensional limitations. It is useful only when theconstraint function is simple enough for one variable to be expressed as a function of the other. TheLagrange multiplier method, on the other hand, can be used to solve problems involving many variablesand more complicated functions.

7.1.2 The Lagrange Multiplier Method

Given the problem max(min)f(x, y) s.t. g(x, y) = b

The Lagrangean Function, or simply the Lagrangean, of this problem is defined as

L(λ, x, y) = f(x, y) + λ[b− g(x, y)] (38)

In this case, λ which is a scalar is called the Lagrange Multiplier. The Lagrangean may also bewritten as

L(λ, x, y) = f(x, y)− λ[g(x, y)− b]

Given the Lagrangean function and your choice variables x, y, and λ, we can now derive the first-order necessary conditions for a local optimal solution.

L(λ, x, y) = f(x, y) + λ[b− g(x, y)]

Lλ =∂L

∂λ= b− g(x, y) = 0

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Lx =∂L

∂x=∂f(x, y)

∂x− λ∂g(x, y)

∂x= 0

Ly =∂L

∂y=∂f(x, y)

∂y− λ∂g(x, y)

∂y= 0

Take note that the 2nd and 3rd first-order conditions yield

∂f(x,y)∂x

∂g(x,y)∂x

= λ and∂f(x,y)∂y

∂g(x,y)∂y

= λ

Equating both together and cross-multiplying, we get

∂f(x,y)∂x

∂g(x,y)∂x

=

∂f(x,y)∂y

∂g(x,y)∂y

→∂f(x,y)∂x

∂f(x,y)∂y

=∂g(x,y)∂x

∂g(x,y)∂y

→ fxfy

=gxgy

Thus, the intuition behind the second and third first-order conditions

gxgy→ −fx

fy= −gx

gy

The condition above states that to maximize or minimize the objective function, it is necessarythat the slope of the curve of the objective function is equal to the slope of the curve of the constraint.In other words, it is necessary that the curve of the objective function is tangent to the curve of theconstraint. The intuition behind the 1st first-order condition states that to maximize or minimize theobjective function, it is also necessary that the points are feasible, i.e. they should lie on the graphof the constraint itself.

To help you better understand the aforementioned intuitions, a graphical argument is presentedbelow. The constrained maximization problem is

max f(x, y) s.t. g(x, y) = b

According to the 1st first-order condition, the feasible points are those that lie on the curve definedby the constraint g(x, y) = b. This is what we call the constraint curve. Consider also the level curvesof the objective function

Figure 31: Illustrating a Constrained Optimization

For convenience, assume that the direction of increasing values of f is upward to the right as indicatedin the graph. Maximizing f(x, y) subject to the constraint g(x, y) = b is finding the highest level curveof f that intersects the constraint curve. This is achieved at the point p∗ = (x∗, y∗), where the level

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curve f(x, y) is tangent to the constraint curve g(x, y) = b. At p∗, the slopes of these two curves areequal.

−fxfy

= −gxgy

This is what the 2nd and 3rd first order conditions imply, Taken together, the 3 FOCs obtainedfrom the Lagrangean function indicate that to maximize or minimize the objective function.

• The points must be feasible, i.e. the constraint must be satisfied.

• The slope of the curve of the objective function must be equal to the slope of the constraint curve.

Once we solve for the optimal x∗, y∗, and λ∗, we have to check whether they are indeed maximizers orminimizers using the second order sufficient conditions. This involves using a bordered Hessian matrix,i.e. a matrix of second-order partial derivatives used in constrained optimization problems.

7.1.3 The Bordered Hessian Matrix

The general form of the bordered Hessian matrix for a constrained optimization problem with twovariables is

H =

∂Lλ∂λ

∂Lλ∂x

∂Lλ∂y

∂Lx∂λ

∂Lx∂x

∂Lx∂y

∂Ly∂λ

∂Ly∂x

∂Ly∂y

=

Lλλ Lλx LλyLxλ Lxx LxyLyλ Lyx Lyy

Assuming that the first-order necessary conditions have been satisfied, the second-order sufficient

conditions are as follows

• If the determinant of H is < 0, then (x∗, y∗) is a strict local minimizer of the objective function.

• If the determinant ofH is > 0, then (x∗, y∗) is a strict local maximizer of the objective function.

Take note that the second-order conditions for a local maximum/minimum using a bordered Hessianmatrix is different from the second-order conditions for a local maximum/minimum using an ordinaryHessian matrix.

For example, consider the optimization problem minx2 + y2 s.t. x+ y = 2

1. Set up the LagrangeanL(λ, x, y) = x2 + y2 + λ(2− x− y)

2. Solve for the three first-order conditions and set each of them equal to zero.

Lλ = 2− x− y = 0

Lx = 2x− λ = 0

Ly = 2y − λ = 0

3. Solve for x∗, y∗, and λ∗. You can do this by first isolating λ from the second and third FOCs,equating both then solving for x or y.

λ = 2x = 2y

x = y

Afterwards, substitute the resulting function into the first FOC (i.e. the constraint).

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2− x− y = 0

2− x− x = 0

2− 2x = 0

x = 1

Since x = y, then y∗ = 1. Lastly, λ∗ = 2x = 2(1) = 2. Thus, the stationary point is

(λ∗, x∗, y∗) = (2, 1, 1)

4. Solve for the bordered Hessian matrix and evaluate its determinant at (λ∗, x∗, y∗) = (2, 1, 1)

H =

0 −1 −1−1 2 0−1 0 2

The determinant of the above Hessian matrix is −4 < 0. Thus, the point (x∗, y∗) = (1, 1) is indeeda local maximizer.

Suppose that the consumer with income m has a utility function U(x1, x2) = x0.751 x0.25

2 wherex1 and x2 are the quantities of the two goods consumed whose prices are p1 and p2 respectively.The consumer’s problem is to maximize his utility subject to his budget constraint. Formulatethe Lagrangean function and find the utility-maximizing quantities x∗1 and x∗2.

7.2 Optimization with One Equality Constraint: The n variable case

Optimization with one equality constraint involving more than two variables is just a straightforwardextension of the two-variable case and may be stated as follows

max(min)f(x1, x2, ..., xn) s.t. g(x1, x2, ..., xn) = b

For convenience, let x = (x1, x2, ..., xn). The Lagrangean function of this problem is

L(λ, x) = f(x) + λ[b− g(x)]

The vector of first-order necessary conditions is

∇(λ, x) =

b− g(x)

f1(x)− λg1(x)f2(x)− λg2(x)

...fn(x)− λgn(x)

= 0

The Hessian matrix is

H(λ, x) =

∂Lλ∂λ

∂Lλ∂x1

. . . ∂Lλ∂xn

∂Lx1∂λ

∂Lx1∂x1

. . .∂Lx1∂xn

......

. . ....

∂Lxn∂λ

∂Lxn∂x1

. . . ∂Lxn∂xn

By the First Order Necessary Condition, given max(min)f(x1, x2, ..., xn) s.t. g(x1, x2, ..., xn) = b

where f and g have continuous first order partial derivatives, we have ∇L(λ, x) = 0. This yields(λ∗, x) = 0 as a stationary point of L.

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By the Second Order Sufficient Condition,

• If the determinant of HL3 is < 0, the determinant of HL4 is < 0,...,determinant of HLn is < 0,then x∗ is a local minimizer.

• If the determinant of HL3 is > 0, the determinant of HL4 is > 0,...,determinant of HLn is > 0,then x∗ is a local maximizer.

For example, consider the optimization problem minx2 + y2 + z2 s.t. x+ y + z = 3

1. Set up the Lagrangean

L(λ, x, y) = x2 + y2 + z2 + λ(3− x− y − z)

2. Solve for the three first-order conditions and set each of them equal to zero.

Lλ = 3− x− y − z = 0

Lx = 2x− λ = 0

Ly = 2y − λ = 0

Lz = 2z − λ = 0

3. Solve for x∗, y∗, z∗, and λ∗. You can do this by first isolating λ from the second and third FOCs,equating both then solving for x or y.

λ = 2x = 2y = 2z

x = y = z

Therefore, we get the stationary point

(λ∗, x∗, y∗, z∗) = (2, 1, 1, 1)

4. Solve for the bordered Hessian matrix and evaluate its determinant at (λ∗, x∗, y∗, z∗) = (2, 1, 1, 1)

H =

0 −1 −1 −1−1 2 0 0−1 0 2 0−1 0 0 2

Now, the determinant of HL3 is −4 < 0 and the determinant of HL4 is −12 < 0. Hence,(x∗, y∗, z∗) = (1, 1, 1) is indeed a local minimizer.

Use the Lagrange multiplier method to solve the problem of finding three positive numbers whosesum is 27 and whose product is a maximum.

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8 Course Problem Sets

All of your grade for the course is hinged on the course problem sets. Each problem set constitutesan equal proportion of your final grade. The deadline for any and all problem sets is on Week 14 ofthe Term. However, it is recommended that students submit the problem sets as early as they can toavoid cramming. Try and space out your work flow by answering the problem set after the lecture orduring your fourth hour. All problem sets may be done computerized using MS Word with EquationEditor/MathType, Pages, or LATEX. Alternatively, you may opt to answer all problems manually andscan your answers. Please ensure that your handwritting is readable if this is the alternative you wouldprefer. All problem sets are to be submitted to their respective Canvas assignment entries on or beforethe Friday of the 14th week of the term. As a safety net, you can also submit your problem set [email protected]. All problem sets should be submitted in PDF form.

In terms of grading, your score will be in the grading scheme of DLSU, that is; 4.0, 3.5, 3.0, 2.5, 2.0,1.5, 1.0, and 0.0. To obtain your final grade, the average of the eight (8) problem sets will be computedfor. Note that the rounding off points will be discussed to you in class.

FREQUENTLY ASKED QUESTIONS

1. Q: Will you accept answers without any solutions?A: No! We will strictly follow a no-solution, no-credit policy

2. Q: Does the Final Answer need to be simplified?A: Yes! Simplify to the furthest you can reasonably without causing rounding errors.

3. Q: Can I collaborate with my fellow classmates?A: Yes! But if there is any evidence of copy-paste answers, such as directly copying thesolutions, final answer, and reasoning of your classmate, the grade will be divided by thesuspected number of copies. You are still in graduate school, try and accomplish thesethings with integrity.

4. Q: Are all problem sets the same in difficulty on the average?A: No! There are some that are harder than others and some that are relatively easy.

5. Q: What are your consultation hours?A: Thursday 19:30 - 20:30. Please set any consultation through my email at least 36 hoursin advance.

6. Q: Do you have any recommended schedule for submission of Problem Sets?A: Yes! Please find the table below.

Problem Set RecommendedSubmission

1 Week 32 Week 53 Week 64 Week 85 Week 96 Week 117 Week 128 Week 14

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8.1 Problem Set 1: Single Variable Differentiation

1. Using the difference quotient, find the derivative of the following functions

• f(x) = 120/x

• f(x) = 90x4 + 60x2

• f(x) =√x

2. Find the derivative or y′

• y = x3+2x+1x2+3

• y = xe

ex

• y = 2− x4e−x

• y =(

x1+x

)5

• y = 2x−1−x−2

3x−1−4x−2

• y = αβx2

• y = (2x3 + 5)3/2

• y =6√x5 +

4√x3 +

√x5 + 2

• y = 1x2− 4

2x2

• y = 1ln(ln(x))−1

• y = eex

• y = x4e−2x

• y =

√√√√x2 − 1

• y =

√x+

√x+√x

• y = 1−√

1+xx

• y = 1(x2+x+1)5

3. Find the first and second-order derivatives of the following functions

• y = 2x− 5

• y = 13x

9

• y = ex ln(x)

• y = ln(ln(x2))

• y = (1 + x2)10

• y = x4

4 + x3

3 + 52

2

• y = 12(ex − e−x)

4. Solve for f ′(x) or dydx using implicit differentiation

• x3 + 3x2y − 6xy2 + 2y3 = 0

• x3 + y3 = 4

• exy = e4x − e5y

• xy + y4 = 4 + 2x

• x = 3 +√x2 + y2

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5. A manufacturer’s cost function is given by C(x) = 0.5x3 − 10x2 + 110x , where x denotes theoutput level. What is the marginal cost when output is 10? Interpret your answer, then verify bycomputing for the actual cost of producing the 11th item.

6. The total revenue obtained from the sale of x cellphones is given by R(x) = 10x− 0.02x2 , whereR(x) denotes revenue and x denotes the quantity sold. Find the marginal revenue when 100cellphones are sold, then interpret your result.

7. A study in transport economics uses the relation T = 0.4K1.06, where K is expenditure onbuilding roads and T is a measure of traffic volume. Find the elasticity of T with respect to K,then interpret the result.

8. The demand for a commodity is given by Q = 100−8P , where P denotes the price and Q denotesthe quantity demanded.

• Find the price elasticity of demand function.

• What is the price elasticity of demand when P = 5? Is the demand price elastic, unit elastic,or inelastic? Interpret the result.

9. The inverse demand function for a commodity is given by P = 100 − 0.025Q , where P denotesthe price and Q denotes the quantity demanded.

• Find the total revenue function.

• Find the marginal revenue when Q = 100. Interpret the result.

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8.2 Problem Set 2: Multivariate Differentiation

1. Find the first-order partial derivatives with respect to all variables of the following functions.

• f(x, y, z) = z3 − 3x2y + 6xyz

• f(x, y, z) = xyzx+y+z

• f(x, y) = exy ln(xy)

• f(x, y) = ex2+xy

• f(x, y) = x3ey2

• f(x, y) = x2

y + y2

x

2. Find the first and second order (both cross and direct) of the following functions. Check to see ifYoung’s Theorem is satisfied.

• f(x, y) = x7 − y7

• f(x, y) = x5 ln(y)

• f(x, y) = (x2 − 2y2)5

• f(x, y) = 5x4y2 − 2xy5

• f(x, y) = x2−4y3

2xy

• f(x, y) = x−yx+y

• f(x, y) = x/y

• f(x, y) = xy2 − exy

3. Prove that if z = (ax+ by)2, then xfx + yfy = 2z.

4. Let z = 12 ln(x2 + y2). Show that ∂2z

∂x2+ ∂2z

∂y2= 0

5. Find the total differential of the following functions.

• f(x, y) = 3x3(8x− 7y)

• m(x, y) = 2xyx+y

• z = (x− 3y)3

• u(x, y) = −5x3 − 12xy − 6y5

• f(x, y) = 7x2y3

• z = ex2−y2

• h(x, y) = x3y + x2y2 + xy3

• f(x, y) =√

9− x2 − y2

6. Find both the linear and quadratic Taylor approximations of the following functions:

• f(x) = xe2x about x0 = 0

• f(x) = AKα about K0 = 0

• f(x) = 5(ln(1 + x)−√

1 + x) about x0 = 0

7. Identify, if any, the local maxima and local minima of the given functions using either the first-derivative test or the second derivative test. In addition, identify the inflection point/s, if any,and determine the regions of concavity and convexity.

• f(x) = 13x

3 − x2 + x+ 10

• m(x) = −2x2 + 8x+ 25

• z(x) = x2

x2+2

• u(x) = 3x5 − 5x3

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8.3 Problem Set 3: Economic Applications of Differentiation

1. Consider a typical consumer whose preferences can be represented by a Constant Elasticity ofSubstitution (CES) utility function U(x1, x2) = (xρ1 + xρ2)−1/ρ where x1 represents the amount ofgood 1 consumed and x2 represents the amount of good 2 consumed.

• Take the total differential of U(x1, xx)

• Prove that the consumer is non-satiated. (Hint: Obtain the marginal utility of each goodand prove that this is positive while stating the relevant domain).

• Prove the phenomenon of diminishing marginal utility, that is, the additional utility peradditional consumption will be lower for every additional consumption unit thereafter.

• Prove that given the relevant domain, the consumer will never be "less" happy with anadditional consumption of any good.

• Demonstrate that Young’s Theorem holds

• Do the first five bullets but using the function U(x1, x2) = (x1 − β1)α1(x2 − β2)α2 whereβ > 0 is some subsistence parameter and 0 ≤ α ≤ 1 is some parameter.

• Do the first five bullets but using the function U(x1, x2) = xα1x1−α2 where 0 ≤ α ≤ 1 is some

parameter.

2. A firm has a cost function C = 13Q

3 − 7Q2 + 111Q+ 50 and a demand function Q = 100− P .

• Write out the total revenue function TR of the firm as some function of Q

• Formulate the profit function π as a function of Q.

• Find the profit-maximizing level of output Q∗

• How much is the maximum profit?

3. A firm’s production function is Q = 12L2 − (1/20)L3, where L denotes the number of workers.

• What size of the workforce (L∗) maximizes the output Q?

• What size of the workforce maximizes output per worker Q/L?

4. A firm produces Q = 2√L units of a commodity when L units of labor are employed. If the

price obtained per unit is 160 pesos, and the price per unit of labor is 40 pesos, what value of Lmaximizes profits?

5. Let R(Q) = 80Q and C(Q) = Q2 + 10Q+ 90. The firm can at most produce 50 units.

• How many units must be produced for the firm to make a profit?

• How many units must be produced for the firm to maximize profits?

6. Mr. Constantine wishes to mark out a rectangular flower bed using a wall of his house as one sideof the rectangle. The other three sides are to be marked by wire netting, of which he has only64 ft. available. What are the length L and width W of the rectangle that would give him thelargest possible planting area? Check to make sure that your answer gives you the largest, notthe smallest area.

7. Consider a general two good utility function U = U(x1, x2)

• Take the total differential of this utility function

• Set dU = 0 and rearrange the result such that the marginal utilities are in ratios in one sideand the slopes in another. What does this condition imply?

• What happens to the condition in the second bullet when x1 is increased? Decreased?

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8.4 Problem Set 4: Techniques of Integration

1. Evaluate the following indefinite integrals using whatever technique is feasible

•∫

(20x2 + 2x−1 − 3x5/3) dx

•∫

1t(ln(t))2

dt

•∫r(ln r)2 dr

•∫

x3

(2−x2)5/2dx

•∫

18x2+124x3+8x

dx

•∫x3ex

2 dx

•∫

5xe5x2+3 dx

•∫

xx2+5x+6

dx

•∫

x2+2x(x+2)(x−1) dx

2. Evaluate the following definite integrals using whatever technique is feasible

•∫ 2

01

(3+5x)2dx

•∫ 3 ln 3

0 (ex/3 − 3) dx

•∫ 10

0 (1 + 0.4t)e−0.05t dt

•∫ 2

0 (x− 2)e−x/2 dx

•∫ ba (px+ q) dx

•∫ 1−1

12(ex + e−x) dx

•∫ 1

0x√

1−x2 dx

3. Evaluate the following improper integrals using whatever technique is feasible and determinewhether it is convergent or divergent

•∫∞−∞ xe

−cx2 dx

•∫∞

0ex

(1+ex)2dx

•∫ 3

1/31

3√3x−1dt

•∫∞e

1x(lnx)2

dx

•∫ a

0x√

a2−x2 dx

•∫∞

1 3x2e−x3 dx

•∫∞

1e−√x

√x

dx

4. Say a firm has a marginal cost function of MC(x) = 0.003x2− 0.08x+ 100 where x represents thequantity produced. Find the total cost function.

5. Suppose a firm has a marginal revenue function equal to MR(x) = 500− 0.1x where x representsthe quantity produced. Find the Total Revenue function of the firm. From there, calculate thedemand function.

6. Consider a demand function P = 200− 0.2Q and a supply function of P = 20 + 0.1Q.

• Compute for the equilibrium price and quantity.

• Calculate for the social surplus or net market surplus

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7. Consider a demand function Q = 500− 50P and a supply function of Q = 50 + 25P .

• Compute for the equilibrium price and quantity.

• Calculate for the consumer surplus

• Calculate for the producer surplus

• Calculate for the social surplus

88


8.5 Problem Set 5: Deep Dive on Matrix Algebra

1. If A and B are square matrices of order n, prove that, in general:

• (A+B)(A−B) 6= A2 −B2

• (A−B)(A−B) 6= A2 − 2AB +B2

What condition must be satisfied for equality to hold in each of the cases above?

2. Identify whether the following statements are "True" or "False"

• The transpose of a symmetric matrix is always equal to the matrix itself.

• The inverse of a symmetric matrix is also symmetric.

• Given any matrix A, we can always derive from it from a transpose and a determinant.

• If the determinant of a square matrix A is zero, then we can be sure that the equation systemAx = d is nonsingular.

• It is possible for a matrix to be its own inverse.

• The product of two symmetric matrices is symmetric.

• Matrix addition, subtraction, and multiplication all satisfy the associative property.

• Matrix multiplication is not commutative, although matrix addition is commutative.

• Non-singular matrices have no inverses

• Idempotent matrices may not necessarily be square matrices.

• For any square matrix, it is always the case that |AB| = |A||B|.

3. Show that the diagonal matrix below can idempotent only if each diagonal element is either 1 or0. Identify how many different numerical idempotent diagonal matrices of dimension n × n canbe constructed altogether from such a matrix.

a11 0 . . . 00 a22 . . . 0. . . . . . . . . . . .0 0 . . . ann

4. Given the following matrices, perform the following operations

A =

(6 −3 −1−4 3 8

), B =

(2 0 51 4 −1

), C =

3 01 −2−1 2

, D =

(1 −22 0

)

• CA• BCD• 3AC − 2D

• A+B

• BTD

• DTA

5. Let A =

1 0 t2 1 t0 1 1

. For what values of t does A have an inverse?

6. Prove that the matrix A = Ix(xTx)−1xT is an idempotent matrix. Do both x and (xTx) have tobe square matrices for the condition to be satisfied?

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7. Write the following systems of linear equations in matrix form

•

ax+ y + z = b1 − azx+ 2y + z = b2

3x+ 4y = b3 − 7z

•

x+ y + z + t = a

3y + 4t+ 2z + x = b

x+ 4y + 8z = c

z + 2x− t = d

•

Qd1 = 10− 2P1 + P2

Qs1 = −2 + 3P1

Qs1 = Qd1

Qd2 = 15 + P1 − P2

Qs2 = −1 + 2P2

Qs2 = Qd2

8. Compute for the determinant and the inverse of the following matrix

A =

2 6 1 71 0 1 24 5 9 83 0 2 1

9. Given the following system of equations, (1) Express the system of linear equations in matrix

format, (2) Find the determinant of your coefficients matrix, (3) Solve for the solution values viamatrix inversion, and (4) Solve for the solution values using Cramer’s Rule

•x1 − 8x2 + x3 = 4

−x1 + 2x2 + x3 = 2

x1 − x2 + 2x3 = −1

•4x+ 3y − 2z = 1

x+ 2y = 6

3x+ z = 4

•x1 + 2x3 + x3 = 4

x1 − x2 + x3 = 5

2x1 + 3x2 − x3 = 1

•2x1 + 2x2 − x3 = 2

x1 − 3X2 + x3 = 0

3x1 + 4x2 − x3 = 1

90


•x+ 3y − 2z = 1

3x− 2y + 5z = 14

2x− 5y + 3z = 1

•2x1 + 4x2 − 3x3 = 12

3x1 − 5x2 + 2x3 = 13

−x1 + 3x2 + 2x3 = 17

•ax− by = 1

bx+ ay = 2

10. What value of x will make the determinant of the matrix below zero.1− x 2 22 1− x 22 2 1− x

11. Suppose that −XTY + xTXC = [0], where X = n × k and Y = n × 1 vector. Solve this system

for matrix C. Show explicitly all the steps involved, the dimensions of this matrix, and the crucialassumption(s) needed to derive the solution, if any.

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8.6 Problem Set 6: Economic Applications of Matrix Algebra

1. The equilibrium conditions for two related markets (siopao and hakaw) are given by

18ph − ps = 87

−2ph + 36ps = 98

Find the equilibrium price for each market using matrix inversion.

2. Given the IS equation 0.3Y + 100i− 252 = 0 and the LM equation 0.25Y − 200i− 176 = 0. Findthe equilibrium level of income and rate of interest.

3. A simple economy is described by the following model

Y = C + I0 +G0

C = C0 + C1Y

• Interpret the parameters C0 and C1 and identify their domains

• Solve for the equilibrium values of the endogenous variables via matrix inversion

• Verify that your solution is correct using Cramer’s Rule

• Based on your solution, what additional condition on the parameter/s will ensure non-singularity of the coefficients matrix.

4. Given Y = C+ I0 where C = C0 + bY , use matrix inversion to find the equilibrium level of Y andC. Analyze these equilibrium forms and give some economic intuition.

5. Consider the following model

Y = C + I

C = 120 + 0.75Y

I = 180− 80R

Md = 0.2Y − 150R

Ms = 210

Find the equilibrium values Y and R using either matrix inversion or Cramer’s Rule. At equilib-rium, what are the values of C, I, and Md. (Item Has a Bonus of + 0.20 to a Raw Problem SetGrade if answered correctly. The problem is slightly difficult)

6. Consider the national income model below

Y − C(Y )− I(i)−G0 = 0

kY + L(i)−Ms0 = 0

0 <∂C

∂Y< 1;

∂I

∂i< 0;

∂L

∂i< 0; k = positive constant

• Is the first equation in the nature of an equilibrium condition?

• What is the total quantity demanded for money in the model?

• Analyze the comparative statics of the model when money supply changes (monetary policy)and when government expenditure changes (fiscal policy). Make sure to indicate the signsand provide the economic meanings.

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Suppose that while demand for money still depends on Y as specified, it is now no longer affectedby the interest rate

• How should the system of equations above be revised?

• Write the new Jacobian and call it J ′. Is |J ′| numerically (in absolute value) larger or smallerthan |J |?• Find the new comparative static derivatives

• Comparing the new (∂Y/∂G0), what can you conclude about the effectiveness of fiscal policyin the new model where money demand is independent of i?

• Comparing the new (∂Y/∂Ms0), what can you conclude about the effectiveness of monetarypolicy in the new model where money demand is independent of i?

7. The production process of a three-industry economy is summarized by the technology matrixbelow

A =

0.3 0 0.40.14 0.2 0.120.05 0.5 0.2

The final demand for the three industries’ goods are 20, 10, and 30 units, respectively.

• Interpret the third column of the technology matrix, and give the economic interpretation of0.14.

• Write the input-output matrix equation for this model, find the Leontief inverse matrix, andinterpret its first column.

• How much of each good should the economy produce to meet the given final demands?

• Calculate the amount of primary input needed to produce the output level of each industry.What is the total amount of primary input required to produce the output levels of allindustries?

• By how much will the output level of each industry change if final demands for industries 1and 2 increase by 50 and 20, respectively, and final demand for industry 3 decreases by 10?

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8.7 Problem Set 7: Unconstrained Optimization

1. Identify the local maxima and/or minima. Verify using either the first- or second-order derivativetest.

• f(x) = ex/x

• f(x) = 2x3 − 5x24x− 7

• f(x) = 19x

3 − 16x

2 − 23x+ 1

• f(x) = x− 2 ln(x+ 1)

2. Consider the total cost function given by TC(Q) = 5Q3 − 10Q2 + 16Q

• Solve for the level of Q that will minimize average cost. Verify your answer using the first-derivative and second-derivative test.

• Using your first-derivative test results in the first bullet, at which range of output is the slopeof the average function positive? At which range of output is this slope of the average costfunction negative?

• Compute for the minimum average cost

3. Given the total cost function TC = 0.5q − 10q2 + 80q

• Find the average cost AC and the marginal cost MC

• At what level of output is average cost at a minimum? Verify using the second-order derivativetest.

• At what level of output is marginal cost at a minimum? Verify using the second-orderderivative test.

4. Suppose that a monopolist has the following demand function and cost function

P (Q) = 10− Q

1000, C(Q) = 5000 + 2Q

Find the output level that maximizes profits, verify whether your answer is indeed a maximum,and find the maximum profit.

5. A quadratic profit function π(Q) = hQ2 +jQ+k is to be used to reflect the following assumptions.First, if nothing is produced, the profit will be negative (because of fixed costs). Second, the profitfunction is strictly concave. Last, the maximum profit occurs at a positive output level Q∗. Giventhe above assumptions, what will be the restrictions on the parameters h, j, and k?

6. A purely competitive firm has a single variable input L (labor), whose price per unit is W0. Itsfixed inputs cost the firm a total of F dollars. The price of the product is P0 and the output levelproduced is Q(L).

• Write out the production function, revenue function, cost function, and profit function of thefirm.

• What is the first-order condition for profit maximization? Give this condition an economicinterpretation.

• Using your answer above, what is the sufficient condition for profit to be maximized, ratherthan minimized?

7. Suppose that a firm has a revenue function given as R(x, y) = ρx + qy and a total cost functiongiven as C(x, y) = αx2 + βy2

• What are the needed first order conditions for profit maximization?

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• What is the optimal x∗ and y∗?

• What is the maximum profit pi∗?

• Verify that ∂π∗

∂ρ = x∗ and ∂π∗

∂q = y∗

8. Consider a firm that produces two products Q1 and Q2 at two different prices P1 and P2. Thecost function is given by C(Q) = 4Q2

1 + 2Q1Q2 + 4Q22.

• Find the profit function

• What are the optimal quantities of Q1 and Q2 that will maximize profit?

• Verify that the optimal quantities will indeed yield a maximum profit.

9. A competitive firm produces output given by the production function Q = L0.25K0.25 , whereLand K are labor and capital respectively. The firm charges a price of 4 PHP per unit of outputproduced, and faces a price of 2 PHP per unit of labor and a price of 4 PHP per unit of capitalemployed.

• Compute for the profit-maximizing levels of labor and capital that the firm should employ.

• Verify that the values of L and K obtained in the first bullet are indeed profit-maximizing.

• Compute for the maximum profit

10. Suppose demand for good 1 is given by p1 = 26 − 3q12 and demand for good 2 is given by p2 =

72−2q2. The firm has the cost function C = 120 + q2 for both goods where q = q1 + q2. Computethe following

• The profit maximizing level of output for each good, i.e., q1 and q2.

• Verification that the quantities obtained in the first bullet will indeed maximize profit

• The prices p1 and p2 that the firm will charge for each good to maximize profit

• The maximum profit

11. Adrienne Inc. faces a Cobb-Douglas production function given by Y = K0.25L0.75, where K isthe number of tools and L is the number of workers. Suppose that each tool costs P100, and thatevery labor unit costs P200. The firm decides to price each name-stamped item at P50. The firmwould like to know how much capital and labor to employ to maximize profits.

• Find the optimal values of capital K and labor L to maximize profits.

• Test whether these will indeed yield maximum profits.

• Find the maximum profit, π∗

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8.8 Problem Set 8: Constrained Optimization

1. A competitive firm charges a price of 50 PHP per output produced, given by Q = L0.25K0.25 .The input costs of labor and capital are 2 PHP and 2 PHP respectively. However, the firm has arequired input capacity of only L+K = 20.

• Compute for the profit-maximizing levels of labor and capital using the Lagrangean method

• Verify that the second-order sufficient condition for a local maxima is met

• Compute for the maximum profit

2. Your preferences are represented by the utility function U = (x0.51 +x0.5

2 )2, where the price of good1 is 4 PHP and the price of good 2 is 3 PHP. Your income is given by 50 PHP.

• Set up the Lagrangean function for this utility-maximization problem with constraint.

• Compute for the utility-maximizing quantities of x1 and x2.

• What is the maximum level of utility you can attain given your utility function and budgetconstraint?

3. Suppose you are trying to minimize your expenditures E = 4x1 + 3x2 subject to the targetmaximum utility level (x0.5

1 + x0.52 )2 = 175

6 .

• Set up the Lagrangean function for this expenditure-minimization problem with constraint

• Compute for the expenditure minimizing quantities of x1 and x2.

• What is the minimum expenditure you can incur given the level of utility you have to attain?

4. Chino has a quasi-linear utility function given by U(x1, x2) = x1 + 6x232 , where prices of goods 1

and 2 are given by p1 and p2, respectively. If Chino’s budget is given by m,

• Solve for the utility-maximizing quantities of x1, x2, and λ

• Find the maximum utility.

• Show that the change in maximum utility due to a peso change in income m is equal to theLagrange multiplier you got in the first bullet.

5. A firm pays a price w for each unit it employs of its labor x. It also incurs fixed costs equal toF . If the firm receives a price p for each unit of output produced, and it produces output givenby f(x) = 4x1/4, how much of labor x should it employ to maximize profits? Verify whether youranswer is indeed a maximum.

6. A firm has three factories each producing the same item. Let x, y and z denote the respectiveoutput quantities that the three factories produce to fulfill an order of 2000 units in total. Thecost functions for the three factories are the following:

C1(x) = 200 +1

100x2

C2(y) = 200 + y +1

300y2 + y

C3(z) = 200 + 10z

• Find the values of x, y and z that minimize the total cost C.

• Verify that the solutions yield minimum costs.

• What is the minimum cost?

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7. A single-product firm intends to produce 30 units of output as cheaply as possible. By using Kunits of capital and L units of labor, it can produce

√K + L units. Suppose that the prices of

labor are 1 and 20, respectively.

• Form the Lagrangean to solve for a minimum cost• Find the optimal choices of K and L• What is the associated cost?• What is the additional cost of producing 31 units rather than 30 units?

8. Each week, an individual consumes x and y of two goods, and works for l hours. These threequantities are chosen to maximize the utility function U(x, y, l) = α ln(x) + β ln(y) + (1 − α −β) ln(L− 1), which is defined for 0 ≤ l < L and for x, y > 0, and where α, β > 0 and α + β < 1.The individual faces the budget constraint px+ qy = wl, where w is the wage per hour.

• State the optimization problem• Solve completely for the utility-maximizing levels of x, y, and l.

9. Alan likes eating chocolates and oranges. His utility is described by U(x1, x2) = (x1 − γ)αxβ2 ,where x1 is chocolate and x2 is oranges. Chocolates are addictive, so γ represents a referencepoint, any amount of chocolates consumed less than γ does not increase utility. The price of x1 isp1 and the price of x2 is p2. Alan’s income is m.

• State the optimization problem.• Find the utility maximizing values of chocolates, x1, and oranges, x2.• Test whether these will indeed yield maximum utility.• Determine how x∗1 changes as γ changes. Why do you think this is so?

10. Consider a consumer who has a direct utility function given as U(x1, x2) = (x0.51 + x0.5

2 )2 wherex1 > 0 and x2 > 0. The consumer chooses to spend his entire income m on the consumption oftwo goods, where each good has a price of p1 and p2 respectively.

• Obtain the marginal utility of each good. Given the domain of the quantities, are thesepositive.• Obtain the marginal rate of substitution for goods 1 and 2 (i.e. MRS12)• Calculate the direct second-order partial derivatives for goods 1 and 2. What economic

principle does this adhere to?• Calculate the cross second-order partial derivatives for goods 1 and 2. Prove that Young’s

Theorem holds.• Use the Lagrangean method to compute for the Marshallian Demand Functions• Generate a Bordered Hessian to determine that the Marshallian demand functions are indeed

going to yield a maximum utility.• Demonstrate that the Marshallian demand functions are homogeneous of degree zero in prices

and income.• Demonstrate that the Marshallian demand functions are strictly increasing in income, strictly

decreasing in prices, and satisfy Walras’ Law.• Obtain the Indirect Utility Function V (p,m) which is a function of prices and income• Demonstrate that the Indirect Utility Function is homogeneous of degree zero in prices and

income.• Demonstrate that the Indirect Utility Function is strictly increasing in income, strictly de-

creasing in prices, and satisfies Roy’s Identity.• Answer the previous bullets but using the direct utility function is U(x1, x2) = 3 ln(x1) +

ln(x2).• Answer the previous bullets but using the direct utility function is U(x1, x2) = −1

x1+ −1

x2.

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9 Course Syllabus

ECO501M: Mathematical Economics for Master in Applied Economics

Instructor: Justin Raymond S. EloriagaClass schedule: Saturday 09:00 - 12:00 (Fully Online) Term 1, A.Y. 2020 - 2021Consultation: By email ([email protected])

9.1 Course Description

This course serves as an introductory course in mathematics for economic analysis at the grad-uate level. The course focuses on the mathematical foundations used in economic theory, andthe objective is for students to learn how to use the necessary mathematical tools in studyingand understanding economics. The course discusses concepts on the applications of differentialcalculus and integral calculus, linear and non-linear optimization, and matrix algebra. At thislevel, it is important that students should be able to successfully complete all of the calculationsneeded with consistency and accuracy, and consequently, develop the ability to interpret andunderstand mathematical equations and calculations. After building on students’ mathematicalfoundations, the course shifts over to economic applications and analyses. At this point, math-ematical theories with economic applications will be covered in class to help students use thelanguage of mathematics to describe and analyze economic models and solve economic problems.

School of Economics’ Course Learning Outcomes:

Knowledge

· Apply both qualitative and quantitative con-cepts of the derivative of a function.· Interpret the concept of a definite integral asthe area of a given region.· Differentiate differential and integral calculusand the relationship between them.

Skills

· Correctly apply differentiation rules.· Apply differential calculus in an economic con-text.· Demonstrate the applicability of integral calcu-lus in the capital accumulation and welfare con-cept of economics.· Solve problems of integration using the differenttechniques of integral calculus.· Solve matrix algebra problems and apply theconcepts in economic applications such as input-output models· Have a full grasp of both linear and non-linearoptimization techniques

Behavior

· Confidently express graphical and conceptualmodels in equation form.· Exhibit resilience in solving economic problemsmathematically.· Exhibit willingness to work well within a team,to be open-minded and receptive to others’ in-sights and constructive feedback, and to developinitiative

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9.2 Main References

9.2.1 Required References

• Chiang, A. and K. Wainwright. (2005). Fundamental Methods of Mathematical Eco-nomics. 4th edition. McGraw-Hill/Irwin: New York. (Main Reference)

• Sydsæter, K. and P. Hammond. (2012). Essential Mathematics for Economic Analysis,4th edition. Pearson Education Limited: England

• Eloriaga, J. (2020). Mathematical Economics

9.2.2 Other References

• Danao, R. (2011). Mathematical Methods in Economics and Business. The University ofthe Philippines Press: Quezon City.

• Danao, R. (2017). Core Concepts of Calculus with Applications. The University of thePhilippines Press: Quezon City.

9.3 Grading System and Requirements

Given the unique circumstance of the term, the subject shall no longer require students toundertake a summative midterm or final examination. Instead, the course material shall bebroken down into eight (8) problem sets divided equally. These 8 problem sets are long problemsets which cover all grounds on the topics planned to be covered. Students are required toaccomplish 8 problem sets and the average of all will be the student’s corresponding grade.All problem sets are individually accomplished but collaboration (to the extent of discussion) iscertainly permitted.

Grading Component 96 ≤ grade ≤ 100 4.0

Long Problem Sets (8 in Total) 100% 90 ≤ grade ≤ 95.9. . . 3.584 ≤ grade ≤ 89.9. . . 3.078 ≤ grade ≤ 83.9. . . 2.572 ≤ grade ≤ 77.9. . . 2.066 ≤ grade ≤ 71.9. . . 1.5

Total 100% 60 ≤ grade ≤ 65.9. . . 1.0NOTE: Passing Mark 72% grade <60 0.0

9.4 Lectures

The course will revolve heavily on the prepared content in the Mathematical Economics lecturenotes by Justin Eloriaga. The class will operate around the content of the material and exam-ples contained inside the lecture notes shall be answered in class. Students are encouraged toimmediately work on the problem sets thereafter to have a consistent flow. In addition, lecturevideos on YouTube (Channel Name: Justin Eloriaga) concerning various topics in MathematicalEconomics are also there for reference. With the class’ consent, all synchronous classes will berecorded and uploaded to an unlisted2 playlist on the same channel as well.

During synchronous lectures, it is highly encouraged that students’ webcam are turned onwhile mics are turned off. No penalty will be incurred from the negligence to follow the statedpolicy. Attendance will not be monitored due to conditions arising from the pandemic.

2Only people with the video link may find and view the video

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9.5 Course Plan

Topics ActivitiesCourse Introduction (Week 1)

Introductory Concepts on DifferentiationThe Difference Quotient LectureThe Derivative Class ExerciseSingle Variable Differentiation Problem Sets

Further Dive on Differentiation (Week 2-3)Multiple Variable Differentiation LectureSingle Variable Optimization Class ExerciseTaylor Approximations Problem SetsImplicit DifferentiationFermat’s TheoremConcavity and ConvexityRolle’s TheoremWeirstrauss’ Theorem

Introductory Integration (Week 4)The Anti-derivative LectureIndefinite Integration Class ExerciseThe Area Under a Curve Problem SetsThe Definite IntegralThe Fundamental Theorem of Calculus

Techniques of Integration (Week 5-6)Integration by Substitution LectureIntegration by Parts Class ExerciseIntegration by Partial Fractions Problem SetsIntegration by Partial Fractions with Repeated FactorsImproper IntegralsArea Between CurvesEconomic Applications of Integration

Introduction to Matrix Algebra (Weeks 7-8)Introduction to Matrices LectureMatrix Operations Class ExerciseSolving Systems of Linear Equations using Matrices Problem SetsDeterminantsCramer’s Rule and Matrix Inversion

Deep Dive on Matrix Algebra (Week 9)Eigenvalues and Eigendecomposition LectureComparative Statics Class ExerciseEconomic Applications of Matrix Algebra Problem Sets

Unconstrained and Constrained Optimization (Weeks 10-11)

The Extremum LectureSolving for Local Maxima and Local Minima Class ExerciseThe Hessian Matrix Problem SetsThe Lagrangean MethodThe Bordered Hessian MatrixLinear ProgrammingThe Lagrangean MethodDuality Conditions

Individual Consultation for Select Topics (Weeks 12-13)Submission of Problem Sets (Week 14)

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9.6 Course Guidelines

1. We shall utilize AnimoSpace as the virtual classroom for the course. All announcementsand classroom materials shall be posted in the Canvas website, which you can access viadlsu.instructure.com. You’ll have to use your DLSU email to log in.

2. We shall devote at most 3 hours of synchronous learning sessions each week, which willcover lectures, class exercises, and live online consultations and discussions. At most 3hours per week shall be allocated to asynchronous learning that shall cover the time forproblem sets and consultations which can be done through AnimoSpace or social mediasuch as Zoom or Facebook.

3. Synchronous learning sessions will be saved for the benefit of the students who cannot beonline on schedule. In addition, short video clips of the topics and concepts covered in eachchapter/module shall be available for the students to catch up. These short videos can beaccessed through the instructor’s YouTube page.

4. As noted above, occasional exercises will be done during synchronous (i.e. conference)learning sessions. The problems are typically included in the slides. Students are encour-aged to collaborate with the rest of the class and ask questions during the session.

5. Problem sets will be assigned after each topic. Asynchronous channels are also availablefor students in case they need assistance with regards to the lessons. The students are alsoencouraged to collaborate with their classmates in answering the problem sets.

6. Given the unique circumstance this term, Week 10 and 13 have been declared as breaks bythe VCA. As such, we will have no synchronous lectures during these weeks. Asynchronousvideos for viewing can be found on my channel regarding these topics.

9.7 Contact and Consultation Hours

My consultation hours are from 18:00 - 19:00 (Thursday) over Zoom. Please set an appointmentat least 24 hours in advance. Consultation is strictly by appointment only. All contact may bemade through [email protected] or through 09260321823. Alternatively, students mayfill up the contact form in justineloriaga.com

Syllabus prepared by:

Justin Raymond S. Eloriaga

Noted by:

Dr. Arlene B. Inocencio Dr. Marites M. TiongcoDepartment Chair Dean

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About Your Professor

Justin Raymond S. Eloriaga is a lecturer at the De La Salle University School of Economics and aCentral Bank Associate at the Bangko Sentral ng Pilipinas. He has taught various courses in both theundegraduate and graduate school with a primary focus on advanced econometrics and microeconomictheory. He obtained his Bachelor of Science in Applied Economics from De La Salle University andgraduated summa cum laude. He also obtained a Master of Science in Economics from the same university,completing the rigorous ladderized track in the fastest time recorded in the school’s history. He was also the17th Commissioner of the Young Economist’s Convention of the Economics Organization and a foundingexecutive board member of the Lasallian Graduate Economics Society. He was also one of the top youngeconomists of 2019 as inducted by the Philippine Economics Society.

Email: [email protected]: justineloriaga.comOffice: School of Economics Office. St. La Salle Hall Room 221, Mezzanine Area

De La Salle UniversitySchool of Economics

2020-2021

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