Calculus Applied to Business Part05B

Calculus Applied to Business and EconomicsFurther Applications of Differential Calculus

Prof. Kenneth James T. Nuguid

May 7, 2013

Prof. Kenneth James T. Nuguid Calculus Applied to Business and Economics

Using First Derivatives to Find Maximum andMinimum Values

Increasing and Decreasing Functions

A function f is increasing over an interval I = (a, b) if, forevery x1 and x2 in I,

if x1 < x2, then f(x1) < f(x2).

(If the input x1 is less than the input x2, then the output for x1is less than the output for x2.)A function f is decreasing over an interval I = (a, b) if, forevery x1 and x2 in I,

if x1 > x2, then f(x1) > f(x2).

(If the input x1 is greater than the input x2, then the outputfor x1 is greater than the output for x2.)



Figure : Intervals of increase and decrease



Theorem 1

If f (x) > 0 for all x in an open interval I = (a, b), then f(x) isincreasing over I.

If f (x) < 0 for all x in an open interval I = (a, b), then f(x) isdecreasing over I.

If the graph of a function f(x) has tangent lines with onlypositive slopes on the interval a < x < b, then the graph will berising and f(x) will be increasing on the interval. Since theslope of each such tangent line is given by the derivative f (x),it follows that f(x) is increasing (rising graph) on intervalswhere f (x) > 0. Similarly, f(x) is decreasing (falling graph) onintervals where f (x) < 0.



Figure : Derivative criteria for increasing and decreasing functions.



Procedure for Using the Derivative to Determine Intervals ofIncrease and Decrease for a Function f

1 Find all values of x for which f (x) = 0 or f (x) isdiscontinuous, and mark these numbers on a number line.This divides the line into a number of open intervals.

2 SIGN TEST: Choose a test number c from each intervala < x < b determined in step 1 and evaluate f (c). Then,

If f (c) > 0, the function f(x) is increasing (rising graph)for a < x < b.If f (c) < 0, the function f(x) is decreasing (falling graph)for a < x < b.



Example 1. Find the intervals of increase and decrease for thefunction

f(x) = 2x3 + 3x2 12x 7.Solution: The derivative of f(x) is

f (x) = 6x2 + 6x 12 = 6(x + 2)(x 1),

which has no discontinuities but f (x) = 0 when x = 1 orx = 2. The numbers 2 and 1 divide the xaxis into threeopen intervals; namely x < 2,2 < x < 1, and x > 1.test number interval f (c) conclusion

c = 3 x < 2 f (3) = 24 > 0 f(x) is increasingc = 0 2 < x < 1 f (0) = 12 < 0 f(x) is decreasingc = 2 x > 1 f (2) = 24 > 0 f(x) is increasing

Exercise 1. Do the same for f(x) = 12 + 9x 3x2 x3.Prof. Kenneth James T. Nuguid Calculus Applied to Business and Economics


Critical Value

A critical value of a function f(x) is any number c in thedomain of f for which the tangent line at (c, f(c)) is horizontalor for which the derivative does not exist. That is, c is a criticalvalue if f(c) exists and

f (c) = 0 or f (c) does not exist.

Relative Extrema

Let I be an open interval in the domain of f(x).

i. f(c) is a relative minimum in an open interval I = (a, b)containing c if f(c) f(x), for all x I;

ii. f(c) is a relative maximum in an open interval I = (a, b)containing c if f(c) f(x), for all x I.



Figure : Relative extrema (minimum or maximum) occur at the peaksand valleys of a graph. They are not necessarily the absolute extrema.



Theorem 2

Relative extrema can only occur at critical points.

To find relative extrema, we need only consider thoseinputs(x-values for which the derivative is 0 or for which thederivative does not exist. We can think of a critical value as acandidate for a value where a relative extremum might occur.



Example 2. Find all critical numbers of the functionf(x) = 2x 4x2 + 3.Solution:The polynomial f(x) is defined for all x, and its derivative is

f (x) = 8x3 8x = 8x(x2 1) = 8x(x 1)(x + 1).

Since the derivative exists for all x, the only critical numbersare where f (x) = 0; that is, x = 0, x = 1, and x = 1.

Exercise 2. Findthe critical numbers of R(t) =63t t2t2 + 63

.



Consider the graph of a function f(x) below.

Remarks: In the graph of f(x) above:

1 c1, c2, c4, c7, and c8 are critical values because f(c) = 0.

2 c3, c5, and c6 are critical values because f(c) does not

exist.Prof. Kenneth James T. Nuguid Calculus Applied to Business and Economics


Figure : Three critical points (c, f(c)) where f (c) = 0. It isimportant to note that while relative extrema occur at critical points,not all critical points correspond to relative extrema.



Figure : Three critical points (c, f(c)) where f (c) does not exist. It isimportant to note that while relative extrema occur at critical points,not all critical points correspond to relative extrema.



How can we tell when the existence of a critical value leads usto a relative extremum? The following graph leads us to a test.



Theorem 3: The First Derivative Test

For any continuous function f that has exactly one critical valuec in an open interval (a, b):

i. f has a relative minimum at c if f (x) < 0 on (a, c) andf (x) > 0 on (c, b) . That is, f is decreasing to the left of cand increasing to the right of c.

ii. f has a relative maximum at c if f (x) > 0 on (a, c) andf (x) < 0 on (c, b) . That is, f is increasing to the left of cand decreasing to the right of c.

iii. f has neither a relative maximum nor a relative minimumat c if f (x) has the same sign on (a, c) as on (c, b).



Using the First Derivative Test

1 Find all critical values by determining where f (x) is 0 andwhere f (x) is undefined (but f(x) is defined). Find f(x)for each critical value.

2 Use the critical values to divide the x-axis into intervalsand choose a test value in each interval.

3 Find the sign of f (x) for each test value chosen in step 2,and use this information to determine where f(x) isincreasing or decreasing and to classify any extrema asrelative maxima or minima.

4 Plot some additional points and sketch the graph.



Example 3. Find the relative extrema of

f(x) = 2x3 3x2 12x + 12.

Solution: We begin by finding a general expression for thederivative:

f (x) = 6x2 6x 12 = 6(x2 x 2) = 6(x + 1)(x 2).

Hence, f (x) = 0 when x = 1 or x = 2. (Critical Values)test no. interval f (c) conclusion2 x < 1 f (2) = 24 > 0 f(x) is increasing

x = 1 f(1) = 19 is a rel. max0 1 < x < 2 f (0) = 12 < 0 f(x) is decreasing

x = 2 f(2) = 8 is a rel. min3 x > 2 f (2) = 60 > 0 f(x) is increasing



Exercise 3. Graph the function g given by g(x) = x3 27x 6and find any relative extrema.



Application of Finding Relative Extrema

Example 4. According to the National Statistics Office, thenumber of professional services employees fluctuated during theperiod 20002009, as modeled by

E(t) = 28.31t3 + 381.86t2 1162.07t + 16, 905.87,

where t is the number of years since 2000 (t = 0 corresponds to2000) and E is in thousands of employees. Find the relativeextrema of this function, and interpret the meaning of therelative extrema.

Solution: Find the derivative of the given funciton:

E(t) = 84.93t2 + 763.72t 1162.07E(t) = 0 when t 1.94 or t 7.05.



Hence, the critical values are t 1.94 or t 7.05.test no. (k) interval sign of E(k) conclusion

1 t < 1.941 E(t) is decreasingt = 1.94 E(1.94) = 15, 881.92

is a rel. min

2 1.94 < t < 7.05 + E(t) is increasing

t = 7.05 E(7.05) = 17, 772.78is a rel. max

8 x > 7.05 E(t) is decreasing



Applications of Finding Relative Extrema

Exercise 4a. Brody Electronics estimates that it will sell Nunits of a new toy after spending a thousands of dollars onadvertising, where

N(a) = a2 + 300a + 6, 0 a 300.

Find the relative extrema of this function, and interpret themeaning of the relative extrema.



Applications of Finding Relative Extrema

Exercise 4b. The revenue derived from the sale of a new kind ofmotorized skateboard t weeks after its introduction is given by

R(t) =63t t2t2 + 63

, 0 t 63.

million dollars. When does maximum revenue occur? What isthe maximum revenue?


Using Second Derivatives to Find Maximum andMinimum Values

Concavity

Suppose that f is a function whose derivative f exists at everypoint in an open interval I = (a, b). Then the graph of f(x) is

i. concave upward on (a, b) if f (x) is increasing on (a, b);ii. concave downward on (a, b) if f (x) is decreasing on

(a, b).



Figure : Another diagram to illustrate concavity



Theorem 4

i. If f(x) > 0 on an interval I, then the graph of f is concaveup. (f (x) is increasing, so f is turning up on I.)

ii. If f(x) > 0 on an interval I, then the graph of f is concavedown. (f (x) is decreasing, so f is turning up on I.)

The Second Derivative Test

For any continuous function f that has exactly one criticalvalue c in an open interval (a, b):

i. f(c) is a relative minimum if f(x) > 0 on (a, b).

ii. f(c) is a relative maximum if f(x) < 0 on (a, b).

For f(c) = 0, the First-Derivative Test can be used todetermine whether f(c) is a relative extremum.



Figure : (L) f(c) is a relative minimum (R) f(c) is a relativemaximum.



Second Derivative Procedure for Determining Intervals ofConcavity for a Function f

1 Find all values of x for which f(x) = 0 or f(x) does notexist, and mark these numbers on a number line. Thisdivides the line into a number of open intervals.

2 Choose a test number c from each interval determined instep 1 and evaluate f(c). Then,

If f(c) > 0, then the graph is concave upward on thisinterval.If f(c) < 0, then the graph is concave downward on thisinterval.

Example 5. Determine intervals of concavity for the function

f(x) = 2x6 5x4 + 7x 3.



Solution: We find that

f (x) = 12x5 20x3 + 7f(x) = 60x4 60x2 = 60x2(x2 1) = 60x2(x + 1)(x 1).

f(x) is continuous at any value of x, Hence, we only need tofind x for which it is 0.f(x) = 0 for x = 0, x = 1, and x = 1. These numbers dividethe x-axis into 4 intervals: namelyx < 1,1 < x < 0, 0 < x < 1, and x > 1.test no. (k) interval Sign of f(k) conclusion

2 x < 1 + f(x) is concave up0.5 1 < x < 0 f(x) is concave down0.5 0 < x < 1 f(x) is concave down2 x > 2 + f(x) is concave up

Exercise 5. Determine the concavity of f(x) = x3 3x + 2.Prof. Kenneth James T. Nuguid Calculus Applied to Business and Economics


Figure : Graph of f(x) = 2x6 5x4 + 7x 3



Steps in Using the 2nd Derivative for Finding Relative Extrema

i. Find f (x) and f (x).ii. Find critical values c of f (x).iii. Find the values f(c) for each critical value c.

If f (c) > 0, then f(c) is a relative minimum.If f (c) < 0, then f(c) is a relative maximum.



Example 6. Find the relative extrema of the function f given by

f(x) = x3 + 3x2 9x 13.Solution:

f (x) = 3x2 + 6x 9 = 3(x + 3)(x 1)f(x) = 6x + 6 = 6(x + 1)

f (x) = 0 when x = 3 or x = 1. So,f(3) = (3)3 + 3(3)2 9(3) 13 = 14f(1) = (1)3 + 3(1)2 9(1) 13 = 18

We use the 2nd derivative to test the critical values -3 and 1:

f(3) = 6(3 + 1) = 12 < 0 f(3) = 14 is a rel. max.f(1) = 6(1 + 1) = 12 > 0 f(1) = 18 is a rel. min.

Exer 6. Find the relative extrema of f(x) = x3 3x2 9x 1.Prof. Kenneth James T. Nuguid Calculus Applied to Business and Economics

Using Derivatives to Find Absolute Maximum andMinimum

An extremum may be at the highest or lowest point for afunctions entire graph, in which case it is called an absoluteextremum. For example, the parabola given by f(x) = x2 hasa relative minimum at (0, 0). This is also the lowest point forthe entire graph of f , so it is also called the absolute minimum.

Figure : Graph of a function with relative extrema at x = c1, c2, c3and absolute minimum at x = c1 while absolute maximum is at x = b.



Extreme Value Theorem

A continuous function f(x) defined on an interval [a, b] musthave an absolute maximum and an absolute minimum on [a, b].

Maximum-Minimum Principle 1

To find the absolute maximum and minimum values of acontinuous function f(x) on a closed interval [a, b]:

i. Find f (x).ii. Determine all critical values in [a, b].

iii. Make a table of values as follows:

a c1 c2 cn bf(a) f(c1) f(c2) f(cn) f(b)

The largest value of f(x) is the absolute max while thesmallest is the absolute min.



Example 7. Find the absolute maximum and minimum values of

f(x) = x3 3x + 2 on[ 2, 3

2

].

Solution: f (x) = 3x2 3. Find the critical values. Thederivative exists for all real numbers. Thus, we merely solvef (x) = 0.

3x2 3 = 03x2 = 3

x2 = 1

x = 1x -2 -1 1 3/2

f(x) 0 4 0 7/8



Hence, absolute maximum = 4 at x = 1 and the absoluteminimum = 0 at x = 2 and x = 1.

Figure : Graph shows absolute max and absolute min

Exercise 7. Do the same but over the interval [3,3/2].Prof. Kenneth James T. Nuguid Calculus Applied to Business and Economics


Maximum-Minimum Principle 2

Suppose the f(x) is a function such that f (x) exists for every xin an interval I and that there is exactly one (critical) value c inI, for which f (c) = 0. Then

i. f(c) is the absolute max over I if f(c) < 0; or

ii. f(c) is the absolute min over I if f(c) > 0.

If f (c) = 0, then either we must use MaximumMinimumPrinciple 1 or we must know more about the behavior of thefunction over the given interval.

Example 8. Find the absolute maximum and minimum values of

f(x) = 4x x2.



Solution: When no interval is specified, we consider the entiredomain of the function. In this case, the domain is the set of allreal numbers.

f (x) = 4 2x.whose critical value is x = 2. Since there is only one criticalvalue, we can apply MaximumMinimum Principle 2 using thesecond derivative:

f(x) = 2.The second derivative is constant. Thus, f(2) = 2 and sincethis is negative, we have the absolute maximum:

f(2) = 4 2 22 = 8 4 = 4.Exercise 8. Find the absolute maximum and minimum of

f(x) = x2 4x.Prof. Kenneth James T. Nuguid Calculus Applied to Business and Economics

Applications: Optimization Problems

Depreciation

Example 9. Suppose that the value V of the inventory at FidosPet Supply decreases, or depreciates, with time t, in months,where

V (t) = 50 25t2

(t + 2)2.

i. Find V (0), V (5), V (10), and V (70).

ii. Find the maximum value of the inventory over the interval[0,].

iii. Does there seem to be a value V (t) below which will neverfall? Explain.



Solution:

V (0) = 50

V (5) = 37.24

V (10) = 32.64

V (70) = 26.37.

V (t) = 25(t + 2)2(2t) t2(2(t + 2))

(t + 2)4

= 252t3 8t2 + 8t 2t3 4t2

(t + 2)4= 254t

2 8t(t + 2)4

= 254t(t + 2)(t + 2)4

= 25 4t(t + 2)3



On the interval [0,], the derivative is continuous. Hence, weonly need to find t for which V (t) = 0.

V (t) = 0 when t = 0.

Finding the value of V (t) for the endpoints of the interval andfor critical values of t, we see that

V (0) = 50 is the absolute maximum.

Exercise 9. The total-cost and total-revenue functions forproducing x items are

C(x) = 5000 + 600x and R(x) = 12x2 + 1, 000x,

where 0 x 600. Use these functions to:i. Find the number of items, for which the total profit is a

maximum.

ii. Find the number of items, for which the average profit is amaximum.


Implicit Differentiation

If an equation has variables x and y and y is not isolated on oneside of the equation, the derivative dy/dxcan be found withoutsolving for y by the method of implicit differentiation. Wedo this because often, it is difficult or impossible to solve for yand to express dydx solely in terms of x. For example, theequation

y3 + x2y5 x4 = 27determines y as a function of x, but it would be difficult tosolve for y.


We can use this formula for the derivative of y without solvingfor y by computing

d

dxyn = nyn1 dy

dx= (nyn1) y.



To differentiate implicitly:

i Differentiate both sides of the equation with respect to x (orwhatever variable you are differentiating with respect to).

ii Apply the rules for differentiation (the Power, Product,Quotient, and Chain Rules) as necessary. Any time anexpression involving y is differentiated, dydx will be a factor inthe result.

iii Isolate all terms with dydx as a factor on one side of theequation.

iv If necessary, factor out dydx .

v If necessary, divide both sides of the equation to isolate dydx .



Example 10. For y3 + x2y5 x4 = 27, find dydx using implicitdifferentiation.

d

dx

(y3 + x2y5 x4 = 27) = d

dx(27)

d

dxy3 +

d

dxx2y5 d

dxx4 = 0

3dy

dxy2 + x2 5y4 dy

dx+ y5 2x 4x3 = 0

3dy

dxy2 + x2 5y4 dy

dx= 4x3 y5 2x(

3y2 + x2 5y4) dydx

= 4x3 y5 2xdy

dx=

4x3 y5 2x3y2 + x2 5y4

Exercise 10. For the demand equation x =

200 p3differentiate implicitly to find dpdx .


Related Rates

A related rate occurs when the rate of change of one variable(with respect to time) can be calculated in terms of the rate ofchange (with respect to time) of another variable of which it isa function.Steps in solving problems of Related Rates:

1 Write the equation that relates the two variables.

2 Differentiate this equation with respect to time t.

3 Substitute the known values of the variables and givenrates.


Related Rates

Business: Rates of Change of Revenue, Cost, and Profit

Example 11. For Kens Landscaping, the total revenue from theyard maintenance of x homes is given by

R(x) = 1, 000x x2,

and the total cost is given by

C(x) = 3, 000 + 20x.

Suppose that Ken is adding 10 homes per day at the momentwhen the 400th customer is signed. At that moment, what isthe rate of change of (a) total revenue, (b) total cost, and (c)total profit?


Related Rates

Solution:

a. dRdt = 1, 000 dxdt 2xdxdt = 1, 000 10 2(400)(10) =2, 000 per day.

b. dCdt = 20 dxdt = 20 10 = 200 per day.c. Since P = R C

dP

dt=dR

dt dC

dt= 2, 000 per day 200 per day= 1, 800 per day.

Exercise 11. Do the same as the previous example given that

R(x) = 50x 0.5x2 and C(x) = 10x + 3,

when x = 10, and dxdt = 5 units per day.


Calculus Applied to Business Part05B

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