Z00 REND1011 11 SE MOD6 PP2 - Pearson Educationwps.pearsoned.co.uk/wps/media/objects/12987/... · M6-6 MODULE 6 • CALCULUS-BASED OPTIMIZATION For example, 4. (M6-7) Note that is

M6-1

1. Find the slope of a curve at any point.

2. Find derivatives for several common types offunctions.

3. Find the maximum and minimum points on curves.

4. Use derivatives to maximize total revenue and otherfunctions.

After completing this module, students will be able to:

6

LEARNING OBJECTIVES

Calculus-Based Optimization

MODULE

Summary • Glossary • Key Equations • Solved Problem • Self-Test • Discussion Questions and

Problems • Bibliography

M6.1 Introduction

M6.2 Slope of a Straight Line

M6.3 Slope of a Nonlinear Function

M6.4 Some Common Derivatives

M6.5 Maximum and Minimum

M6.6 Applications

MODULE OUTLINE

Z00_REND1011_11_SE_MOD6 PP2.QXD 2/21/11 12:48 PM Page 1

M6-2 MODULE 6 • CALCULUS-BASED OPTIMIZATION

M6.1 Introduction

Although most of the quantitative techniques presented in this book are based on algebra, thereare many situations in which calculus and derivatives are helpful in finding the best solution toa business problem. In this module, we begin by reviewing the slope of a line, as this is impor-tant in understanding the derivative of a function. Then we explain what a derivative is and howit is used.

M6.2 Slope of a Straight Line

Recall that the equation of a line can be written as

(M6-1)

where b is the slope of the line.If we are given any two points, and that are on a line, the slope is

(M6-2)

where (delta) is used to represent “change in.”For example, the slope of the line that passes through the points (2,3) and (4,7) is

We can find the intercept, a, by using this slope and either point in the equation for a line. Forthis example, if we select the point (2, 3), we have

The equation of this line is then

This is illustrated in Figure M6.1. Notice that the slope of this line (and any other straight line)is constant. We can pick any two points on the line and compute the slope using Equation M6-2.For a nonlinear equation, the slope is not constant, as we explain in the next section.

Y = -1 + 2X

a = -1

3 = a + 2122 Y = a + bX

b =

¢Y

¢X=

7 - 3

4 - 2=

4

2= 2

¢

b =

Change in Y

Change in X=

¢Y

¢X=

Y2 - Y1

X2 - X1

1X2, Y22,1X1, Y12Y = a + bX

The equation of a straight line isY = a + bX.

The slope of the line is thechange in Y over the change in X.

⎧⎪⎨⎪⎩

⎧⎪⎪⎨⎪⎪⎩

Y = –1 + 2X

Y

X

�Y = 7 – 3 = 4

�X = 4 – 2 = 2

20

2

–2

4

–4

6

8

10

12

14

–2 4 6 8

(2,3)

(4,7)

FIGURE M6.1Graph of Straight Line


Y = X 2 – 4X + 6

–2 0 2 4 6

12

10

8

6

4

2

Y

X

FIGURE M6.2Graph of QuadraticFunction

M6.3 SLOPE OF A NONLINEAR FUNCTION M6-3

M6.3 Slope of a Nonlinear Function

Figure M6.2 provides a graph of the function

Notice that this is not a straight line. To determine the slope of a curve at any point, we find theslope of a line tangent to the curve at this point. For example, if we wish to find the slope of theline when we find a line that is tangent at that point. To find the slope of a line, we useEquation M6-2, which requires us to have two points. We want the slope at so we let

and we find

Thus, one point will be (3,3). We wish to find another point close to this, so we will choose avalue of close to If we pick for the other point (an arbitrary choice used sim-ply to illustrate this process), we find

This other point is (5,11).The slope of the line between these two points is

To get a better estimate of the slope where we choose a value of even closer toTaking the point (3,3) with the point (4,6), we have

Neither of these slopes would be exactly the same as the slope of the line tangent to this curve atthe point (3,3), as shown in Figure M6.3. However, we see that they each give us an estimate ofthe slope of a tangent line. If we keep selecting points closer and closer to the point where

we find slopes that are closer to the value we are trying to find.To get a point very close to (3,3), we will use a value, and add this to to get the

second point. As we let get smaller, we find a point closer to the originalpoint (3,3). From the original equation,

we have

Y1 = 32- 4132 + 6 = 3

Y = X2- 4X + 6

1X + ¢X, Y22¢XX = 3¢X,

X = 3,

b =

¢Y

¢X=

6 - 3

4 - 3=

3

1= 3

X1 = 3.X2X1 = 3,

b =

¢Y

¢X=

11 - 3

5 - 3=

8

2= 4

Y2 = 1522 - 4152 + 6 = 11

X2 = 5X1 = 3.X2

Y1 = (3)2- 4(3) + 6 = 3

X1 = 3X = 3,

X = 3,

Y = X2- 4X + 6

If a curve is nonlinear, we canfind the slope at any point byfinding the slope of a tangent line at that point.



and

Thus,

The slope is then

As gets smaller, the value of approaches the slope of the tangent line. The value ofthe slope (b) in this example will approach 2. This is called the limit as approaches zero andis written as

It is obvious that at other points on the curve, the tangent line would have a different slope as Yand would not be the values given here.

To find a general expression for the slope of the tangent line at any point on a curve, we canrepeat this process for a general point X. Let and and let and represent the corresponding values for Y. For an equation of the form

we let

and

Y2 = a(X + ¢X)2+ b(X + ¢X) + c

Y1 = aX2+ bX + c

Y = aX2+ bX + c

Y2Y1X2 = X + ¢X,X1 = X

¢Y

lim¢X:0

1¢X + 22 = 2

¢X¢Y>¢X¢X

b =

¢Y

¢X=

¢X2+ 2¢X

¢X=

¢X1¢X + 22¢X

= ¢X + 2

¢Y = Y2 - Y1 = 1¢X2+ 2¢X + 32 - 3 = ¢X2

+ 2¢X

= 19 + 6¢X + ¢X22 - 12 - 4¢X + 6 = ¢X2+ 2¢X + 3

Y2 = 13 + ¢X22 - 413 + ¢X2 + 6

The limit as X approaches 0 is used to find the slope of atangent line.

¢

4 6 82–2 0

5

–5

–10

–15

10

15

20

Tangent line at (3,3)

Line through (3,3) and (4,6)

Line through (3,3) and (5,11)

Y = X 2 – 4X + 6

X

Y

FIGURE M6.3Graph of Tangent Lineand Other LinesConnecting Points


M6.4 SOME COMMON DERIVATIVES M6-5

The derivative of a function isused to find the slope of thecurve at a particular point.

M6.4 Some Common Derivatives

Although derivatives exist for a number of different functional forms, we restrict our discussionto the six most common ones. The references at the end of this module provide additionalinformation on derivatives. Table M6.1 gives a summary of the common derivatives. We provideexamples of each of these.

1. (M6-4)

For example, if then The graph of Y is a horizontal line, so the change in Y iszero, regardless of the value of X.

2. (M6-5)

For example,

3. (M6-6)If Y = cXn, then Y¿cnXn-1

if Y = X9, then Y¿ = 9X9-1= 9X8

if Y = X3, then Y¿ = 3X3-1= 3X2

if Y = X2, then Y¿ = 2X2-1= 2X

If Y = Xn, then Y¿ = nXn-1

Y¿ = 0.Y = 4,

c = constant

If Y = c, then Y¿ = 0The derivative of a constant is 0.

Expanding these expressions and simplifying, we find

Then

Taking the limit as X approaches zero, we have

This is the slope of the function at the point X, and it is called the derivative of Y. It is denotedas or The definition of a derivative is

(M6-3)

Fortunately, we will be using some common derivatives that are easy to remember, and it is notnecessary to go through this process of finding the limit every time we wish to find a derivative.

Y¿ =

dY

dX= lim

¢X:0 e ¢Y

¢Xf

dY>dX.Y¿

lim¢X:0

(b + 2aX + c¢X) = b + 2aX

¢

¢Y

¢X=

b(¢X) + 2aX(¢X) + c(¢X)2

¢X=

¢X(b + 2aX + c¢X)

¢X= b + 2aX + c¢X

¢Y = Y2 - Y1 = b(¢X) + 2aX(¢X) + c(¢X)2

FUNCTION DERIVATIVE

Y C Y� 0

Y Xn Y� nXn�1

Y cXn Y� cnXn�1

Y g(x) + h(x) Y� g�(x) + h�(x)

Y g(x) � h(x) Y� g�(x) � h�(x)==

==

Y¿ =

-n

Xn+1Y =

1

Xn

==

==

==

TABLE M6.1Some CommonDerivatives



For example,

4. (M6-7)

Note that is the same as For example,

5. (M6-8)

For example,

6. (M6-9)

For example,

Second DerivativesThe second derivative of a function is the derivative of the first derivative. This is denoted as or For example, if

then the first derivative is

and

The second derivative tells us about the slope of the first derivative, and it is used in finding themaximum and minimum of a function. We use this in the next section.

= 72X2+ 24X

Y¿ ¿ =

d2Y

dX2 = 24(3)X3-1+ 12(2)X2-1

= 24X3+ 12X2

Y¿ =

dY

dX= 6(4)X4-1

+ 4(3)X3-1

Y = 6X4+ 4X3

d2Y>dX2.Y–

if Y = 2X4- 4X2 then Y¿ = 2(4)X4-1

- 4(2)X2-1= 8X3

- 8X

if Y = 5X3- X2 then Y¿ = 5(3)X3-1

- 2X2-1= 15X2

- 2X

If Y = g(x) - h(x), then Y¿ = g¿(x) - h¿(x)

if Y = 5X4+ 3X2 then Y¿ = 5(4)X4-1

+ 3(2)X2-1= 20X3

+ 6X

if Y = 2X3+ X2 then Y¿ = 2(3)X3-1

+ 2X2-1= 6X2

+ 2X

If Y = g(x) + h(x), then Y¿ = g¿(x) + h¿(x)

if Y =

2

X4, then Y¿ = 2(-4)X-4-1=

-8

X5

if Y =

1

X3 (or Y = X-3), then Y¿ = -3X-3-1= -3X-4

=

-3

X4

Y = X-n.Y =

1

Xn,

If Y =

1

xn, then Y¿ = - nX-n-1=

- n

Xn+1

if Y = 2X4, then Y¿ = 2(4)X4-1= 8X3

if Y = 4X3, then Y¿ = 4(3)X3-1= 12X2

If a function is the sum ordifference of two functions, thederivative is the sum ordifference of the individualderivatives of those twofunctions.

The second derivative is thederivative of the first derivative.

M6.5 Maximum and Minimum

In using quantitative techniques in business, we often try to maximize profit or minimize cost. Ifa profit function or cost function can be developed, taking a derivative may help us to find theoptimum solution. In dealing with nonlinear functions, we often look at local optimums, whichrepresent the maximums or minimums within a small range of X.


M6.5 MAXIMUM AND MINIMUM M6-7

4 6 10800 2

Y = X 3 – 4X 2 + 12X + 313A

B

X

YFIGURE M6.4Graph of Curve withLocal Maximum andLocal Minimum

Figure M6.4 illustrates a curve where point A is a local maximum (it is higher than thepoints around it), point B is a local minimum (it is lower than the points around it), but there isno global maximum or minimum as the curve continues to increase without bound as Xincreases, and it decreases without bound as X decreases. If we place limits on the maximumand minimum values for X, then the endpoints can be checked to see if they are higher than anylocal maximum or lower than any local minimum.

To find a local optimum, we take the derivative of the function and set it equal to zero.Remember that the derivative gives the slope of the function. For a point to be a local maximumor minimum, the tangent line must be a horizontal line, which has a slope of zero. Therefore,when we set the derivative equal to zero and solve, we find a value of X that might be a localmaximum or minimum. Such a point is called a critical point.

The following function generated the graph in Figure M6.4:

Point A gives a local maximum and point B gives a local minimum. To find the values of Xwhere these occur, we find the first derivative and set this equal to zero:

Solving this for X, we factor this and have

so the critical points occur when and The second derivative is

At the first critical point, so

Because this is negative, this point is a local maximum. At the second critical point, where

Because this is positive, this critical point is a local minimum.Consider Figure M6.5, which is a graph of

This derivative is equal to zero when The second derivative is

Y– = 3(2)X2-1= 6X

X = 0.

Y¿ = 3X2

Y = X3

Y– = 2(8) - 8 = 8

X = 8,

Y– = 2(2) - 8 = -4

X = 2,

Y– = 2X - 8

X = 6.X = 2

(X - 2)(X - 6) = 0

Y¿ = X2- 8X + 12 = 0

Y =

1

3X3

- 4X2+ 12X + 3

A local maximum (or minimum)is the highest (or lowest) point ina neighborhood around thatpoint.

To find a local optimum, we findthe first derivative, set it equal to0, and solve for X. This point iscalled a critical point.



When Thus, this is neither a maximum nor a minimum but is a point ofinflection, as shown in Figure M6.5.

A critical point will be

1. a maximum if the second derivative is negative.

2. a minimum if the second derivative is positive.

3. a point of inflection if the second derivative is zero.

X = 0, Y– = 6(0) = 0.

M6.6 Applications

There are many problems in which derivatives are used in business. We discuss a few ofthese here.

Economic Order QuantityIn Chapter 6, we show the formula for computing the economic order quantity (EOQ),which minimizes cost when certain conditions are met. The total cost formula under theseconditions is

where

order quantity

annual demand

ordering cost per order

holding cost per unit per year

purchase (material) cost per unit

The variable is Q and all of the others are known constants. The derivative of the total cost withrespect to Q is

dTC

dQ=

-DCo

Q2 +

Ch

2

C =

Ch =

Co =

D =

Q =

TC =

D

Q Co +

Q

2 Ch + DC

Total cost = (Total ordering cost) + (Total holding cost) + (Total purchase cost)

The EOQ model is derived fromthe total inventory cost function.

–1.5 –1

Y = X 3

–0.5 0.50

–0.5

–1

–1.5

0.5

1

1.5

1 1.5

Y

X

FIGURE M6.5Graph of Function withPoint of Inflection atX 0�


M6.6 APPLICATIONS M6-9

Setting this equal to zero and solving results in

We cannot have a negative quantity, so the positive value is the minimum cost. The secondderivative is

If all the costs are positive, this derivative will be positive for any value of . Thus, thispoint must be a minimum.

The formula for the EOQ model demonstrated here is the same formula used in Chapter 6.This method of using derivatives to derive a minimum cost quantity can be used with many totalcost functions, even if the EOQ assumptions are not met.

Total RevenueIn analyzing inventory situations, it is often assumed that whatever quantity is produced can besold at a fixed price. However, we know from economics that demand is a function of the price.When the price is raised, the demand declines. The function that relates the demand to the priceis called a demand function.

Suppose that historical sales data indicate that the demand function for a particularproduct is

where

quantity demanded (or sold)

price in dollars

The total revenue function is

Substituting for Q (using the preceding demand function) in this equation gives us

A graph illustrating this total revenue function is provided in Figure M6.6. To find the price thatwill maximize the total revenue, we find the derivative of total revenue:

Setting this equal to zero and solving, we have

Thus, to maximize total revenue, we set the price at $6. The quantity that will be sold at thisprice is

The total revenue is

TR = PQ = 6(3,000) = $18,000

Q = 6,000 - 500 P = 6,000 - 500(6) = 3,000 units

P = 6

TR¿ = 6,000 - 1,000P

TR = 6,000 P - 500 P2

TR = P(6,000 - 500 P)

TR = PQ

Total revenue = Price * Quantity

P =

Q =

Q = 6,000 - 500P

Q 7 0

d2TC

dQ2 =

DCo

Q3

Q = ;A2DCo

Ch

Total revenue equals totaldemand times the selling price.

To maximize total revenue, findthe derivative of the totalrevenue function, set it equal to0, and solve.



Key Equations

(M6-1)Equation of a straight line.

(M6-2)

Slope of a straight line found from two points.

(M6-3)

Definition of derivative.

(M6-4)Derivative of a constant function.If Y = c where c is any constant, then Y¿ = 0

Y¿ =

dY

dX= lim

¢X:0e ¢Y

¢Xf

b =

Change in Y

Change in X=

¢Y

¢X=

Y2 - Y1

X2 - X1

Y = a + bX (M6-5)Derivative of X raised to a power.

(M6-6)Derivative of a constant times a function of X.

(M6-7)Derivative of to a power.

(M6-8)Derivative of the sum of two functions of X.

(M6-9)Derivative of the difference of two functions of X.If Y = g(x) - h(x), then Y¿ = g¿(x) - h¿(x)

If Y = g(x) + h(x), then Y¿ = g¿(x) + h¿(x)

1>XIf Y = 1>Xn, then Y¿ = -nX-n-1= -n>Xn+1

If Y = cXn, then Y¿cnXn-1

If Y = Xn, then Y¿ = nXn-1

Y = 6,000X – 500X 2

0

2,000

4,000

6,000

8,000

10,000

12,000

14,000

16,000

18,000

20,000

0 5Price

10 15

Rev

enue

FIGURE M6.6Total Revenue Function

Summary

In this module we use the concept of limits to demonstrate thatthe derivative is the slope of a curve. We use several commonderivatives to find the maximums and minimums of functions.The local optimums are found by taking the derivative of afunction, setting it equal to zero, and solving. If the second de-rivative is negative at this point, the local optimum is a maxi-mum. If the second derivative is positive at this point, the local

optimum is a minimum. If the second derivative is zero, thepoint is called a point of inflection.

Derivatives can be used to find the EOQ, which minimizestotal inventory cost. A total revenue function is based on a de-mand curve. The derivative of this shows where revenues aremaximized.

Glossary

Critical Point A point on a curve where the first derivativeis equal to zero.

Derivative The slope of a function at a particular point. It isthe limit of the change in Y over the change in X, as thechange in X approaches 0.

Local Maximum The highest point in the neighborhoodaround that point.

Local Minimum The lowest point in the neighborhoodaround that point.

Point of Inflection A point on a curve where the firstderivative is equal to zero, but the point is neither amaximum nor a minimum.

Second Derivative The derivative of the first derivative.Slope The slope of a line is the change in the Y value divided

by the change in the X value.


SELF-TEST M6-11

Solved Problem M6-1The following function relates the price for a product to the quantity sold:

Find the price that will maximize total revenue.

Solution

So,

Setting this equal to zero and solving, we have

or

Then,

P = 6 - 0.75Q = 6 - 0.75(4) = 6 - 3 = 3

Q = 4

6 - 1.5Q = 0

TR¿ = 6 - 0.75(2)Q

TR = (6 - 0.75Q)Q = 6Q - 0.75Q2

TR = PQ

Total revenue = Price * Quantity

P = 6 - 0.75Q

Solved Problem

� Before taking the self-test, refer back to the learning objectives at the beginning of the module and the glossary at the end ofthe module.

� Use the key at the back of the book to correct your answers.� Restudy pages that correspond to any questions that you answered incorrectly or material you feel uncertain about.

c. the value of the second derivative is positive.d. the value of the first derivative is negative.

5. A point of inflection occurs whena. the first derivative is 0.b. the second derivative is 0.c. the first and the second derivatives are both 0.d. the first and the second derivatives are both positive.

6. The EOQ model can be derived from the derivative ofa. the holding cost function.b. the ordering cost function.c. the purchase cost function.d. the total inventory cost function.

7. To find the quantity that maximizes total revenue, we shoulda. set total revenue equal to total cost.b. find the derivative of the total cost function.c. find the derivative of the total profit function.d. find the derivative of the total revenue function.

1. The slope of a curve at a particular point can be founda. by computing the first derivative of the function for

that curve.b. by computing the second derivative of the function for

that curve.c. by computing the change in X divided by the change

in Y.d. by dividing the value of Y at that point by the value of X.

2. If the first derivative is set equal to zero and solved for X,at that point (X) the curve could bea. at a maximum.b. at a minimum.c. at an inflection point.d. any of the above.

3. A critical point is a point wherea. the slope of the function is 0.b. the Y value for the function is 0.c. the X value for the function is 0.d. the value of is 0.

4. A function is at a maximum at the point At thispoint,a. the value of the second derivative is 0.b. the value of the second derivative is negative.

X = 5.Y/X

Self-Test



BibliographyBarnett, Raymond, Michael Ziegler, and Karl Byleen. College Mathematics

for Business, Economics, Life Sciences, and Social Sciences, 9th ed.Upper Saddle River, NJ: Prentice Hall, 2002.

Haeussler, Ernest, and Richard Paul. Introductory Mathematical Analysis forBusiness, Economics and the Life and Social Sciences, 10th ed. UpperSaddle River, NJ: Prentice Hall, 2002.

Discussion Questions and Problems

Discussion QuestionsM6-1 Explain how to find the slope of a straight line.

M6-2 Explain how to find the slope of a nonlinear functionat a particular point, using limits.

M6-3 What is the procedure for finding the maximum orminimum of a function? How is the second deriva-tive used in this process?

M6-4 What are critical points?

ProblemsM6-5 Find the derivative of the following functions.

(a)(b)(c)(d)

M6-6 Find the second derivative for each of the functionsin Problem M6-5.

M6-7 Find the derivatives of the following functions.(a)(b)(c)(d)

M6-8 Find the second derivative for each of the functionsin Problem M6-7.

M6-9 Find the critical point for the function Is this a maximum, minimum, or point of

inflection?

M6-10 Find the critical points for the function

Identify each criticalpoint as a maximum or a minimum or a point ofinflection.

A13 BX3- 5X2

+ 25X + 12.

Y =

5X + 4.Y = 6X2

-

Y = 25>2X4Y = 2>X3Y = 5X4

+ 12X2+ 10X + 21

Y = X6- 0.5X2

- 16

Y = 25>X4Y = 1>X2Y = 4X5

+ 2X3- 12X + 1

Y = 2X3- 3X2

+ 6

M6-11 Find the critical point for the function Is this a maximum, minimum, or point of inflection?

M6-12 The total revenue function for a particular product isFind the quantity that will

maximize total revenue. What is the total revenuefor this quantity?

M6-13 The daily demand function for the AutoBrightcar washing service has been estimated to be

where Q is the quantity of cars and Pis the price. Determine what price would maximizetotal revenue. How many cars would be washed atthis price?

M6-14 The total revenue function for AutoBright car wash-ing service (see Problem M6-13) is believed to beincorrect due to a changing demand pattern. Basedon new information, the demand function is nowestimated to be Find the price thatwould maximize total revenue.

M6-15 The total cost function for the EOQ model is

In a particular inventory situation, annual demand(D) is 2,000, ordering cost per order is $25,holding cost per unit per year is $10, andpurchase (material) cost per unit (C) is $40. Writethe total cost function with these values. Take thederivative and find the quantity that minimizes cost.

M6-16 Find the second derivative of the total cost functionin Problem M6-15 and verify that the value at thecritical point is a minimum.

(Ch)(Co)

TC =

D

Q Co +

Q

2 Ch + DC

Q = 180 - 2P2.

Q = 75 - 2P,

TR = 1,200Q - 0.25Q2.

Y = X3- 20.


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Z00 REND1011 11 SE MOD6 PP2 - Pearson Educationwps.pearsoned.co.uk/wps/media/objects/12987/... · M6-6 MODULE 6 • CALCULUS-BASED OPTIMIZATION For example, 4. (M6-7) Note that is