Introductory maths analysis chapter 17 official

INTRODUCTORY MATHEMATICAL INTRODUCTORY MATHEMATICAL ANALYSISANALYSISFor Business, Economics, and the Life and Social Sciences

2007 Pearson Education Asia

Chapter 17 Chapter 17 Multivariable CalculusMultivariable Calculus


INTRODUCTORY MATHEMATICAL ANALYSIS

0. Review of Algebra

1. Applications and More Algebra

2. Functions and Graphs

3. Lines, Parabolas, and Systems

4. Exponential and Logarithmic Functions

5. Mathematics of Finance

6. Matrix Algebra

7. Linear Programming

8. Introduction to Probability and Statistics


9. Additional Topics in Probability

10. Limits and Continuity

11. Differentiation

12. Additional Differentiation Topics

13. Curve Sketching

14. Integration

15. Methods and Applications of Integration

16. Continuous Random Variables

17. Multivariable Calculus

INTRODUCTORY MATHEMATICAL ANALYSIS


• To discuss functions of several variables and to compute function values.

• To compute partial derivatives.

• To develop the notions of partial marginal cost, marginal productivity, and competitive and complementary products.

• To find partial derivatives of a function defined implicitly.

• To compute higher-order partial derivatives.

• To find the partial derivatives of a function by using the chain rule.

Chapter 17: Multivariable Calculus

Chapter ObjectivesChapter Objectives


• To apply the second-derivative test for a function of two variables.

• To find critical points for a function.

• To develop the method of least squares.

• To compute double and triple integrals.


Chapter ObjectivesChapter Objectives


Functions of Several Variables

Partial Derivatives

Applications of Partial Derivatives

Implicit Partial Differentiation

Higher-Order Partial Derivatives

Chain Rule

Maxima and Minima for Functions of Two Variables

17.1)

17.2)

17.3)


Chapter OutlineChapter Outline

17.4)

17.5)

17.6)

17.7)


Lagrange Multipliers

Lines of Regression

Multiple Integrals

17.8)

17.9)

17.10)


Chapter OutlineChapter Outline



17.1 Functions of Several Variables17.1 Functions of Several Variables

Example 1 – Functions of Two Variables

• A function can involve 2 or more variables, e.g.

22

2,

yxyxfz

a. is a function of two variables. Because the

denominator is zero when y = 2, the domain of f is all (x, y) such that y 2.

b. h(x, y) = 4x defines h as a function of x and y. The domain is all ordered pairs of real numbers.

c. z2 = x2 + y2 does not define z as a function of x and y.

2

3,

y

xyxf



17.1 Functions of Several Variables

Example 3 – Graphing a Plane

Sketch the plane

Solution:

The plane intersects the x-axis when y = 0 and z = 0.

.632 zyx



17.1 Functions of Several Variables

Example 5 – Sketching a SurfaceSketch the surfaceSolution: z = x2 is a parabola.

.2xz



17.2 Partial Derivatives17.2 Partial Derivatives

• Partial derivative of f with respect to x is given by

• Partial derivative of f with respect to y is given by

h

yxfyhxfyxf

hx

,,lim,

0

h

yxfhyxfyxf

hy

,,lim,

0



17.2 Partial Derivatives

Example 1 – Finding Partial Derivatives

If f(x, y) = xy2 + x2y, find fx(x, y) and fy(x, y). Also, find fx(3, 4) and fy(3, 4).

Solution: The partial derivatives are

Thus, the solutions are

xyyyxyyxfx 221, 22

22 212, xxyxyxyxfy

334 ,3

404 ,3

xy

x

f

f




Example 3 – Partial Derivatives of a Function of Three VariablesIf find

Solution:

By partial differentiating, we get

,,, 322 zzyxzyxf .,, and ,, ,,, zyxfzyxfzyxf zyx

xzyxfx 2,,

yzzyxfy 2,,

22 3,, zyzyxfz




Example 4 – Finding Partial Derivatives

If find

Solution:

By partial differentiating, we get

,,,,22 tsrt

rsuutsrgp

. and ,

1,1,1,0t

p

t

p

s

p

22

2

222

22 2

srtt

srtru

tsrt

strsurutsrt

s

p

222

22222 2

2tsrt

srtrsu

s

psrttsrtrsu

t

p

0

1,1,1,0

t

p



17.3 Applications of Partial Derivatives17.3 Applications of Partial Derivatives

Example 1 – Marginal Costs

• Interpretation of rate of change:

A company manufactures two types of skis, the Lightning and the Alpine models. Suppose the joint-cost function for producing x pairs of the Lightning model and y pairs of the Alpine model per week is

where c is expressed in dollars. Determine the marginal costs ∂c/∂x and ∂c/∂y when x = 100 and y = 50, and interpret the results.

fixed. held is when to respect with of change of rate the is

fixed. held is when to respect with of change of rate the is

xyzy

z

yxzx

z

6000857507.0, 2 yxxyxfc



17.3 Applications of Partial Derivatives

Example 1 – Marginal Costs

Solution: The marginal costs are

Thus,

and

85$

89$7510014.0

50,100

50,100

y

c

x

c

85 and 7514.0

y

cx

x

c




Example 3 – Marginal Productivity

A manufacturer of a popular toy has determined that the production function is P = √(lk), where l is the number of labor-hours per week and k is the capital (expressed in hundreds of dollars per week) required for a weekly production of P gross of the toy. (One gross is 144 units.) Determine the marginal productivity functions, and evaluate them when l = 400 and k = 16. Interpret the results.




Example 3 – Marginal Productivity

Solution: Since ,

Thus, lk

l

k

P

lk

kklk

l

P

2 and

22

1 2/1

2/1lkP

2

5 and

10

1

16,40016,400

klkl k

P

l

P



17.4 Implicit Partial Differentiation17.4 Implicit Partial Differentiation

Example 1 – Implicit Partial Differentiation

• We will look into how to find partial derivatives of a function defined implicitly.

If , evaluate when x = −1, y = 2, and

z = 2.

Solution: Using partial differentiation, we get

022

yyx

xzx

z

0 2

02

0

22

22

zyxx

yz

x

z

xzyxzx

zyxxz

xy

xyx

xz

x

2

2,2,1

x

z



17.5 Higher-Order Partial Derivatives17.5 Higher-Order Partial Derivatives

Example 1 – Second-Order Partial Derivatives

• We obtain second-order partial derivatives of f as

Find the four second-order partial derivatives of

Solution:

yyyyxyyx

yxxyxxxx

f fff

f fff

)(means and )( means

)(means and )( means

., 222 yxyxyxf

xyxyxfyyyxf

xyxyyxf

xyxx

x

42, and 22,

22,2

2

xyxyxfxyxf

yxxyxf

yxyy

y

42, and 2,

2,2

22



17.5 Higher-Order Partial Derivatives

Example 3 – Second-Order Partial Derivative of an Implicit Function

Determine if

Solution: By implicit differentiation,

Differentiating both sides with respect to x, we obtain

Substituting ,

2

2

x

z

.2 xyz

0 2

2

zz

y

x

zxy

xz

x

x

zyz

x

zyz

xx

z

x

2

2

21

2

1

2

1

z

y

x

z

2

0 422

13

22

2

2

z

z

y

z

yyz

x

z



17.6 Chain Rule17.6 Chain Rule• If f, x, and y have continuous partial derivatives,

then z is a function of r and s, and

s

y

y

z

s

x

x

z

s

z

r

y

y

z

r

x

x

z

r

z

and



17.6 Chain Rule

Example 1 – Rate of Change of Cost

For a manufacturer of cameras and film, the total cost c of producing qC cameras and qF units of film is given by

The demand functions for the cameras and film are given by

where pC is the price per camera and pF is the price per unit of film. Find the rate of change of total cost with respect to the price of the camera when pC = 50 and pF = 2.

900015.030 FFC qqqqcc

FCF

F

ppqppc

qc 4002000 and 9000



17.6 Chain Rule

Example 1 – Chain Rule


Solution: By the chain rule,

Thus,

11015.09000

015.0302

C

FC

FC

F

FC

C

CC

qpp

qp

q

p

c

p

q

q

c

p

c

2.1232,50

FC ppCp

c

a. Determine ∂y/∂r ifSolution: By the chain rule,

.3 and 6ln 642 srxxxy

6ln6

2312 4

4

45 x

x

xsrx

r

x

x

y

r

y



17.6 Chain Rule


b. Given that z = exy, x = r − 4s, and y = r − s, find ∂z/∂r in terms of r and s.Solution: By the chain rule,

Since x = r − 4s and y = r − s,

xyeyxr

y

y

z

r

x

x

z

r

z

22 45

452 srsr

sryxrx

esrr

z



17.7 Maxima and Minima for Functions 17.7 Maxima and Minima for Functions of Two Variablesof Two Variables

• Relative maximum at the point (a, b) is shown as

RULE 1

Find relative maximum or minimum when

RULE 2 Second-Derivative Test for Functions of Two Variables

Let D be the function defined by

1. If D(a, b) > 0 and fxx(a, b) < 0, relative maximum at (a, b);

2. If D(a, b) > 0 and fxx(a, b) > 0, relative minimum at (a, b);

3. If D(a, b) < 0, then f has a saddle point at (a, b);

4. If D(a, b) = 0, no conclusion.

yxfbaf ,,

0,

0,

yxf

yxf

y

x

.,,,, 2yxfyxfyxfyxD xyyyxx



17.7 Maxima and Minima for Functions of Two Variables

Example 1 – Finding Critical Points

Find the critical points of the following functions.

a.

Solution:

Since

we solve the system and get

b.Solution: Sincewe solve the system and get and

13522, 22 yxxyyxyxf

,0322, and 0524, yxyxfyxyxf yx

21

1

y

x

lkklklf 33,

,03, and 03, 22 lkklfklklf kl

0

0

k

l . 31

31

k

l




Example 1 – Finding Critical Points

c.

Since

we solve the system and get

1001002,, 22 yxzyxyxzyxf

0100,,

02,,

04,,

yxzyxf

zyxzyxf

zyxzyxf

z

y

x

175

75

25

z

y

x




Example 3 – Applying the Second-Derivative Test

Examine f(x,y) = x3 + y3 − xy for relative maxima or minima by using the second derivative test.Solution: We find critical points,

which gives (0, 0) and (1/3, 1/3).

Now,

Thus,

D(0, 0) < 0 no relative extremum at (0, 0).

D(1/3,1/3)>0 and fxx(1/3,1/3)>0 relative minimum at (1/3,1/3)

Value of the function is

03, and 03, 22 xyyxfyxyxf yx

1, 6, 6, yxfyyxfxyxf xyyyxx

136166, 2 xyyxyxD

271

31

313

313

31

31

31 , f




Example 5 – Finding Relative Extrema

Examine f(x, y) = x4 + (x − y)4 for relative extrema.

Solution: We find critical points at (0,0) through

D(0, 0) = 0 no information.

f has a relative (and absolute) minimum at (0, 0).

0, 12, 02112,

04, 044,222

333

yxfyxyxf yxxyxf

yxyxfyxxyxf

xyyyxx

yx




Example 7 – Profit Maximization

A candy company produces two types of candy, A and B, for which the average costs of production are constant at $2 and $3 per pound, respectively. The quantities qA, qB (in pounds) of A and B that can be sold each week are given by the joint-demand functions

where pA and pB are the selling prices (in dollars per pound) of A and B, respectively. Determine the selling prices that will maximize the company’s profit P.

BAB

ABA

ppq

ppq

29400

400




Example 7 – Profit Maximization

Solution:

The total profit is given by

The profits per pound are pA − 2 and pB − 3,

The solution is pA = 5.5 and pB = 6.

Since ∂2P/∂p2A< 0, we indeed have a maximum.

01342 and 0122

32

BAB

BAA

BBAA

ppp

Ppp

p

P

qpqpP

800 1600 8002

2

2

2

2

ABBA pp

P

p

P

p

P

06 ,5.5 D



17.8 Lagrange Multipliers17.8 Lagrange Multipliers

Example 1 – Method of Lagrange Multipliers

• Lagrange multipliers allow us to obtain critical points.

• The number λ0 is called a Lagrange multiplier.

Find the critical points for z = f(x,y) = 3x − y + 6, subject to the constraint x2 + y2 = 4.Solution: ConstraintConstruct the function

Setting , we solve the equations to be

04, 22 yxyxg

463,,,, 22 yxλyxyxgλyxfλyxF

0 λyx FFF

4

10 and

2

1,

2

3 λ

λy

λx

04

021

023

22 yx

λy

λx



17.8 Lagrange Multipliers

Example 3 – Minimizing CostsSuppose a firm has an order for 200 units of its product and wishes to distribute its manufacture between two of its plants, plant 1 and plant 2. Let q1 and q2 denote the outputs of plants 1 and 2, respectively, and suppose the total-cost function is given by

How should the output be distributed in order to minimize costs?

.2002,, 2221

2121 qqqqλqqfc




Example 3 – Minimizing Costs

Solution: We minimize c = f(q1, q2), given the constraint

q1 + q2 = 200. 2002002,, 21

2221

2121 qqλqqqqλqqF

0200

02

04

21

212

211

qqλ

F

λqqq

F

λqqq

F

150 ,50 21 qq




Example 5 – Least-Cost Input Combination

Find critical points for f(x, y, z) = xy+ yz, subject to the constraints x2 + y2 = 8 and yz = 8.Solution:

Set

We obtain 4 critical points:

(2, 2, 4) (2,−2,−4) (−2, 2, 4) (−2,−2,−4)

88,,,, 222

121 yzλyxλyzxyλλzyxF

08

08

0

02

02

2

1

222

21

1

yzF

yxF

λyyF

λzλyzxF

λxyF

λ

λ

z

y

x

yz

yx

λ

λzλyzx

λx

y

88

1

022

222

21

1



17.9 Lines of Regression17.9 Lines of Regression

2

11

2

1111

2

n

ii

n

ii

n

iii

n

ii

n

ii

n

ii

xxn

yxxyx

a

2

11

2

111

n

ii

n

ii

n

ii

n

iii

n

ii

xxn

yxyxn

b



17.9 Lines of Regression

Example 1 – Determining a Linear-Regression Line

By means of the linear-regression line, use the following table to represent the trend for the index of total U.S. government revenue from 1995 to 2000 (1995 = 100).




Example 1 – Determining a Linear-Regression Line

Solution: Perform the arithmetic

83.921916

7362127486

3.8821916

27482173691

2

2

b

a

xy 83.93.88ˆ



17.10 Multiple Integrals17.10 Multiple Integrals

Example 1 – Evaluating a Double Integral

• Definite integrals of functions of two variables are called (definite) double integrals, which involve integration over a region in the plane.

Find

Solution:

. 121

1

1

0

dxdyxx

3

2

23

22

12

1

1

231

1

10

1

1

1

0

xxx

dxyxy

dxdyx

x

x




Example 3 – Evaluating a Triple Integral

Find

Solution:

. 1

0 0 0

dxdydzxx yx

8

1

82

1

0

41

0 0

22

1

0 0

0

1

0 0 0

xdx

xyyx

dxdyxzdxdydzx

x

xyx

x yx

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