4 Constrained Optimization: The Method of Lagrange Multipliers. Chapter 7 Section 4 Constrained Optimization: The Method of Lagrange Multipliers 551

45.f(x, y)

46.f(x, y) 6x2 12xy y4x 16y 3

47.f(x, y) 2x4 y4 x2(11y18)

48.Sometimes you can classify the critical points of a function by inspecting its level curves. In each case shown in the figure, determine the nature of the critical point of

fat (0, 0).

In many applied problems, a function of two variables is to be optimized subject to a restriction or constrainton the variables. For example, an editor, constrained to stay within a fixed budget of $60,000, may wish to decide how to divide this money between development and promotion in order to maximize the future sales of a new book. If xdenotes the amount of money allocated to development, ythe amount allo-cated to promotion, and f(x, y) the corresponding number of books that will be sold, the editor would like to maximize the sales function f(x, y) subject to the budgetary constraint that x y60,000.

For a geometric interpretation of the process of optimizing a function of two vari-ables subject to a constraint, think of the function itself as a surface in three-dimen-sional space and of the constraint (which is an equation involving xand y) as a curve in the xy plane. When you find the maximum or minimum of the function subject to the given constraint, you are restricting your attention to the portion of the surface that lies directly above the constraint curve. The highest point on this portion of the surface is the constrained maximum, and the lowest point is the constrained mini-mum. The situation is illustrated in Figure 7.20.

y y x x (a) (b) f = 1 f = 2 f = 3 f = 1 f = –3 f = –2 f = –1 f = 3 f = 2 f = 1 f = –3 f = –2 f = –1 f =1 3 f =1 2 LEVEL CURVES x2xy7y2 x ln y

Constrained

Optimization:

The Method

of Lagrange

Multipliers

4

FIGURE 7.20 Constrained and unconstrained extrema.

You have already seen some constrained optimization problems in Chapter 3. (For instance, recall Example 5.1 of Section 3.5.) The technique you used in Chapter 3 to solve such a problem involved reducing it to a problem of a single variable by solv-ing the constraint equation for one of the variables and then substitutsolv-ing the resultsolv-ing expression into the function to be optimized. The success of this technique depended on solving the constraint equation for one of the variables, which is often difficult or even impossible to do in practice. In this section, you will see a more versatile tech-nique called the method of Lagrange multipliers, in which the introduction of a

thirdvariable (the multiplier) enables you to solve constrained optimization problems without first solving the constraint equation for one of the variables.

More specifically, the method of Lagrange multipliers uses the fact that any rel-ative extremum of the function f(x, y) subject to the constraint g(x, y) kmust occur at a critical point (a, b) of the function

F(x, y) f(x, y) [g(x, y) k]

where is a new variable (the Lagrange multiplier). To find the critical points of

F, compute its partial derivatives

Fx fx gx Fy fy gy F (g k)

and solve the equations Fx 0, Fy 0, and F 0 simultaneously, as follows:

Fx fx gx 0 or fx gx

Fy fy gy 0 or fy gy

F (g k) 0 or gk

Finally, evaluate f(a, b) at each critical point (a, b) of F.

y z

x Constraint curve

Constrained maximum Unconstrained maximum

There is a version of the second partials test that can be used to determine what kind of constrained relative extremum corresponds to each critical point (a, b) of F. The techniques required to carry out such an analysis are discussed in more advanced texts, but in this text, we will assume that if fhas a constrained maximum (minimum), it will be given by the largest (smallest) of the critical values f(a, b). Here is a sum-mary of the procedure used in the method of Lagrange multipliers.

A geometric justification of the multiplier method is given at the end of this sec-tion. In the following example, the method is used to solve the problem from Exam-ple 5.1 in Section 5 of Chapter 3.

The highway department is planning to build a picnic area for motorists along a major highway. It is to be rectangular with an area of 5,000 square yards and is to be fenced off on the three sides not adjacent to the highway. What is the least amount of fenc-ing that will be needed to complete the job?

Solution

Label the sides of the picnic area as indicated in Figure 7.21 and let f denote the amount of fencing required. Then,

f(x, y) x2y

The goal is to minimize fsubject to the constraint that the area must be 5,000 square yards; that is, subject to the constraint

xy 5,000

A Procedure for Applying the Method of Lagrange Multipliers

Step 1. Write the problem in the form:

Maximize (minimize) f(x, y) subject to g(x, y) k Step 2. Simultaneously solve the equations

fx(x, y) gx(x, y)

fy(x, y) gy(x, y)

g(x, y) k

Step 3. Evaluate f at all points found in step 2. If the required maximum (minimum) exists, it will be the largest (smallest) of these values.

EXAMPLE 4.1

EXAMPLE 4.1

FIGURE 7.21 Rectangular picnic area.

x

y y

Highway Picnic area

Let g(x, y) xy and use the partial derivatives

fx 1 fy 2 gx y and gy x

to obtain the three Lagrange equations

1 y 2 x and xy 5,000

From the first and second equations you get and (since y 0 and x 0), which implies that

or x 2y

Now substitute x 2yinto the third Lagrange equation to get

2y2 5,000 or y 50

and then use y 50 in the equation x2yto get x 100. It follows that x 100 and y 50 are the values that minimize the function f(x, y) x2ysubject to the constraint that xy 5,000. That is, the optimal picnic area is 100 yards wide (along the highway), extends 50 yards back from the road, and requires 100 50 50 200 yards of fencing.

Find the maximum and minimum values of the function f(x, y) xy subject to the constraint x2 y2 8.

Solution

Let g(x, y) x2y2and use the partial derivatives

fx y fy x gx 2x and gy 2y

to get the three Lagrange equations

y 2x x 2y and x2 y2 8

Neither x nor y can be zero if all three of these equations are to hold (do you see why?), and so you can rewrite the first two equations as

and

which implies that y or x2 y2

x x y 2 x y 2 y x 1 y 2 x 2 x 1 y

EXAMPLE 4.2

EXAMPLE 4.2

Now substitute x2y2into the third equation to get

2x2 8 or x 2

If x 2, it follows from the equation x2 y2that y 2 or y 2. Similarly, if

x 2, it follows that y 2 or y 2. Hence, the four points at which the con-strained extrema can occur are (2, 2), (2, 2), (2, 2), and (2, 2). Since

f(2, 2) f(2, 2) 4 and f(2, 2) f(2, 2) 4

it follows that when x2 y2 8, the maximum value of f(x, y) is 4, which occurs at the points (2, 2) and (2, 2), and the minimum value is 4, which occurs at (2,

2) and (2, 2).

For practice, check these answers by solving the optimization problem using the methods of Chapter 3.

In each of the preceding examples, the first two Lagrange equations were used to eliminate the new variable , and then the resulting expression relat-ing xand ywas substituted into the constraint equation. For most constrained optimization problems you encounter, this particular sequence of steps will often lead quickly to the desired solution.

The method of Lagrange multipliers can be extended to constrained optimization prob-lems involving functions of more than two variables and more than one constraint. For instance, to optimize f(x, y, z) subject to the constraint g(x, y, z) k, you solve

fx gx fx gy fz gz and g k

Here is an example of a problem involving this kind of constrained optimization.

A jewel box is to be constructed of material that costs $1 per square inch for the bot-tom, $2 per square inch for the sides, and $5 per square inch for the top. If the total volume is to be 96 in.3what dimensions will minimize the total cost of construction? Solution

Let the box be x inches deep, yinches long, and zinches wide, as indicated in the accompanying figure. Then the volume of the box is V xyz and the total cost of construction is given by C 1yz 2(2xy 2xz) 5yz 6yz 4xy 4xz abc Top adddegddbgdddddec Sides abc Bottom

L

M

F

T

V

Note

EXAMPLE 4.3

EXAMPLE 4.3

You wish to minimize C6yz 4xy4xz subject to Vxyz96. The Lagrange equations are

Cx Vx or 4y4z (yz)

Cy Vy or 6z 4x (xz)

Cz Vz or 6y 4x (xy)

and xyz 96. Solving each of the first three equations for , you get

By multiplying across each equation, you obtain 4xyz 4xz26yz2 4xyz

4xy24xyz 6y2z 4xyz

6xyz 4x2y 6xyz 4x2z

which can be further simplified by first subtracting the common xyzterms from both sides of each equation

4xz26yz2

4xy2 6y2z

4x2y 4x2z

and then dividing z2from both sides of the first equation, y2from the second, and x2

from the third to get

4x 6y

4x6z so that yz x

4y 4z

Finally, by substituting these expressions into the constraint equation xyz 96, you find that

then y z (6) 4

Thus, the minimal cost occurs when the jewel box is 6 inches deep with a square base, 4 inches on a side.

2 3 x3216 4 9x 3 96 x

In the next two examples, the method of Lagrange multipliers is used to solve con-strained optimization problems from economics.

Autility functionU(x, y) measures the total satisfaction or utilitya consumer receives from having xunits of one particular commodity and yunits of another. The problem is to determine how many units of each commodity the consumer should buy to maximize utility while staying within a fixed budget. The application of the method of Lagrange multipliers to the utility problem is illustrated in the next example.

A consumer has $600 to spend on two commodities, the first of which costs $20 per unit and the second $30 per unit. Suppose that the utility derived by the consumer from xunits of the first commodity and yunits of the second commodity is given by the Cobb-Douglas utility function U(x, y) 10x0.6y0.4. How many units of each commodity should the consumer buy to maximize utility?

Solution

The total cost of buying xunits of the first commodity at $20 per unit and yunits of the second at $30 per unit is 20x30y. Since the consumer has only $600 to spend, the goal is to maximize utility U(x, y) subject to the budgetary constraint that 20x

30y 600.

The three Lagrange equations are

6x0.4y0.4 20 4x0.6y0.6 30 and 20x 30y 600 From the first two equations you get

9x0.4y0.4 4x0.6y0.6

9y 4x or y x

Substituting this into the third equation, you get

from which it follows that

x18 and y (18) 8

That is, to maximize utility, the consumer should buy 18 units of the first commodity and 8 units of the second.

4 9 20x30

A

E

EXAMPLE 4.4

EXAMPLE 4.4

FIGURE 7.22 Budgetary constraint and optimal indifference curve.

Recall from Section 1 that the level curves of a utility function are known as

indifference curves. A graph showing the relationship between the optimal indiffer-ence curve U(x, y) C, where CU(18, 8) and the budgetary constraint 20x30y

600, is sketched in Figure 7.22.

An important class of problems in business and economics involves determining an optimal allocation of resources subject to a constraint on those resources. Here is an example in which sales are maximized subject to a budgetary constraint.

An editor has been allotted $60,000 to spend on the development and promotion of a new book. It is estimated that if xthousand dollars is spent on development and y

thousand on promotion, approximately f(x, y) 20x3/2y copies of the book will be sold. How much money should the editor allocate to development and how much to promotion in order to maximize sales?

Solution

The goal is to maximize the function f(x, y) 20x3/2y subject to the constraint

g(x, y) 60, where g(x, y) x y. The corresponding Lagrange equations are

30x1/2y 20x3/2 and x y 60

ALLOCATION OF RESOURCES

(18, 8) Budget line: 20x + 30y = 600 Optimal indifference curve: U(x, y) = C

20

30 x

y

From the first two equations you get

30x1/2y 20x3/2

Since the maximum value of fclearly does not occur when x0, you may assume that x0 and divide both sides of this equation by 30x1/2 to get

Substituting this expression into the third equation, you get or

from which it follows that

x36 and y (36) 24

That is, to maximize sales, the editor should spend $36,000 on development and $24,000 on promotion. If this is done, approximately f(36, 24) 103,680 copies of the book will be sold.

A graph showing the relationship between the budgetary constraint and the level curve for optimal sales is shown in Figure 7.23.

FIGURE 7.23 Budgetary constraint and optimal sales level.

Budget line: x + y = 60 Optimal sales level: f(x, y) = 103,680

60 60 y x (36, 24) 2 3 5 3x60 x2 3x60 y2 3x

You can solve most constrained optimization problems by the method of Lagrange multipliers without actually obtaining a numerical value for the Lagrange multiplier

. In some problems, however, you may want to compute . This is because has the following useful interpretation.

Suppose the editor in Example 4.5 is allotted $61,000 instead of $60,000 to spend on the development and promotion of the new book. Estimate how the additional $1,000 will affect the maximum sales level.

Solution

In Example 4.5, you solved the three Lagrange equations

30x1/2y 20x3/2 and x y 60

to find that the maximum value M of f(x, y) subject to the constraint x y 60 occurred when x36 and y 24. To find , substitute these values of xand yinto the first or second Lagrange equation. Using the second equation, you get

20(36)3/2 4,320

which means that maximal sales will increase by approximately 4,320 copies (from 103,680 to 108,000) if the budget is increased from $60,000 to $61,000.

Although a rigorous explanation of why the method of Lagrange multipliers works involves advanced ideas beyond the scope of this text, there is a rather simple geo-metric argument that you should find convincing. This argument depends on the fact that for the level curve F(x, y) C, the slope is given by

dy dx Fx Fy

W

M

L

M

W

T

S

L

M

The Lagrange Multiplier ■ Suppose M is the maximum (or

mini-mum) value of f(x, y), subject to the constraint g(x, y) k. The Lagrange mul-tiplier is the rate of change of M with respect to k. That is,

Hence,

change in Mresulting from a 1-unit increase in k dM

dk

This result is true for any level curve of a function F whose partial derivatives exist (provided Fy 0). An example justifying the formula is outlined in Problem 48.

Now, consider the constrained optimization problem: Maximize f(x, y) subject to g(x, y) k

Geometrically, this means you must find the highest level curve of fthat intersects the constraint curve g(x, y) k. As Figure 7.24 suggests, the critical intersection will occur at a point where the constraint curve is tangent to a level curve; that is, where the slope of the constraint curve g(x, y) kis equal to the slope of a level curve f(x, y) C.

FIGURE 7.24 Increasing level curves and the constraint curve.

According to the formula stated at the beginning of this discussion, you have Slope of constraint curve Slope of level curve

or, equivalently,

If you let denote this common ratio, then

and fy gy fx gx fx gx fy gy gx gy fx fy

Direction in which C increases

Point of tangency

Highest level curve,

f(x, y) = C, intersecting

the constraint

Constraint curve: g(x, y) = k

x y

from which you get the first two Lagrange equations

fx gx and fy gy

The third Lagrange equation

g(x, y) k

is simply a statement of the fact that the point of tangency actually lies on the con-straint curve.

In Problems 1 through 16, use the method of Lagrange multipliers to find the indicated extremum. You may assume the extremum exists.

1.Find the maximum value of f(x, y) xysubject to the constraint x y1.

2.Find the maximum and minimum values of the function f(x, y) xy subject to the constraint x2 y2 1.

3.Find the minimum value of the function f(x, y) x2y2subject to the constraint

xy 1.

4.Find the minimum value of the function f(x, y) x2 2y2 xy subject to the constraint 2x y 22.

5.Find the minimum value of f(x, y) x2y2subject to the constraint x2y24.

6.Let f(x, y) 8x2 24xy y2. Find the maximum and minimum values of the function f(x, y) subject to the constraint 8x2 y2 1.

7.Let f(x, y) x2 y22y. Find the maximum and minimum values of the func-tion f(x, y) subject to the constraint x2 y2 1.

8.Find the maximum value of f(x, y) xy2subject to the constraint xy 1.

9.Let f(x, y) 2x24y23xy2x 23y 3. Find the minimum value of the function f(x, y) subject to the constraint x y15.

10.Let f(x, y) 2x2 y2 2xy 4x 2y 7. Find the minimum value of the function f(x, y) subject to the constraint 4x2 4xy 1.

11.Find the maximum and minimum values of f(x, y) exysubject to x2 y2 4.

12.Find the maximum value of f(x, y) ln (xy2) subject to 2x2 3y2 8 for x

0 and y 0.

13.Find the maximum value of f(x, y, z) xyzsubject to x2y 3z 24.

14.Find the maximum and minimum values of f(x, y, z) x 3y z subject to

z2x2 y2.

15.Find the maximum and minimum values of f(x, y, z) x 2y 3zsubject to

x2 y2 z2 16.

P . R . O . B . L . E . M . S

7.4

16.Find the minimum value of f(x, y, z) x2 y2 z2subject to 4x2 2y2 z2 4.

17.A farmer wishes to fence off a rectangular pasture along the bank of a river. The area of the pasture is to be 3,200 square meters, and no fencing is needed along the river bank. Find the dimensions of the pasture that will require the least amount of fencing.

18.There are 320 meters of fencing available to enclose a rectangular field. How should the fencing be used so that the enclosed area is as large as possible?

19.According to postal regulations, the girth plus length of parcels sent by fourth-class mail may not exceed 108 inches. What is the largest possible volume of a rectan-gular parcel with two square sides that can be sent by fourth-class mail?

20.According to the postal regulation given in Problem 19, what is the largest vol-ume of a cylindrical can that can be sent by fourth-class mail? (A cylinder of radius R and length H has volume R2H.)

21.Use the fact that 12 fluid ounces is (approximately) 6.89cubic inches to find the dimensions of the 12-ounce soda can that can be constructed using the least amount of metal. (Recall that the volume of a cylinder of radius rand height his r2h, that the circumference of a circle of radius ris 2r, and that the area of a circle of radius

ris r2.)

22.A cylindrical can is to hold 4cubic inches of frozen orange juice. The cost per square inch of constructing the metal top and bottom is twice the cost per square inch of constructing the cardboard side. What are the dimensions of the least expensive can?

23.A manufacturer has $8,000 to spend on the development and promotion of a new product. It is estimated that if xthousand dollars is spent on development and y

thousand is spent on promotion, sales will be approximately f(x, y) 50x1/2y3/2

units. How much money should the manufacturer allocate to development and how much to promotion to maximize sales?

ALLOCATION OF FUNDS PACKAGING PACKAGING POSTAL PACKAGING PROBLEM 20 PROBLEM 19 x Girth = 2πx y x x y Girth = 4x POSTAL PACKAGING CONSTRUCTION CONSTRUCTION

24.If xthousand dollars is spent on labor and ythousand dollars is spent on equipment, the output at a certain factory will be Q(x, y) 60x1/3y2/3 units. If $120,000 is available, how should this be allocated between labor and equipment to generate the largest possible output?

25.Use the Lagrange multiplier to estimate the change in the maximum output of the factory in Problem 24 that will result if the money available for labor and equipment is increased by $1,000.

26.Recall from Problem 35 of Section 1 that an empirical formula for the surface area of a person’s body is

S(W, H) 0.0072W0.425H0.725

where W (kg) is the person’s weight and H(cm) is his or her height. Suppose for a short period of time, Maria’s weight adjusts as she grows taller so that WH

160. With this constraint, what height and weight will maximize the surface area of Maria’s body?

In Problems 27 and 28, you will need to know that a closed cylinder of radius R

and length Lhas volume V R2L and surface area S 2RL2R2. The volume of a hemisphere of radius R is V R3and its surface area is S 2R2.

27.A bacterium is shaped like a cylindrical rod. If the volume of the bacterium is fixed, what relationship between the radius Rand length Hof the bacterium will result in minimum surface area?

28.A bacterium is shaped like a cylindrical rod with two hemispherical “caps” on the ends. If the volume of the bacterium is fixed, what must be true about its radius R

and length Lto achieve minimum surface area?

29.The thin lens formula in optics says that the focal length Lof a thin lens is related to the object distance doand image distance diby the equation

If Lis fixed, what is the maximum distance s do di between the object and

the image? 1 do 1 di 1 L OPTICS MICROBIOLOGY PROBLEM 28 PROBLEM 27 L Radius R R H MICROBIOLOGY 2 3

SURFACE AREA OF THE HUMAN BODY MARGINAL ANALYSIS ALLOCATION OF FUNDS

30.A jewelry box is constructed by partitioning a box with a square base as shown in the accompanying figure. If the box is designed to have volume 800 cm3, what dimensions should it have to minimize its total surface area (top, bottom, sides, and interior partitions). Notice that we have said nothing about where the partitions are located. Does it matter?

31.Suppose the jewelry box in Problem 30 is designed so that the material in the top costs twice as much as the material in the bottom and sides and three times as much as the material in the interior partitions. Find the dimensions that minimize the total cost of constructing the box.

32.Having disposed of Scélérat’s gunmen (Problem 51 in Section 4, Chapter 6), the spy goes looking for his enemy. He enters a room and the door slams behind him. Immediately, he begins to feel warm, and too late, he realizes he is trapped inside Scélérat’s dreaded broiler room. Searching desperately for a way to survive, he notices that the room is shaped like the circle x2 y260 and that he is stand-ing at the center (0, 0). He presses the stem on his special heat-detectstand-ing wrist-watch and sees that the temperature at each point (x, y) in the room is given by

T(x, y) x2 y23xy 5x15y 130

From an informant’s report, he knows that somewhere in this room there is a trap door leading outside the castle, and he reasons that it must be located at the coolest point in the room. Where is it? Just how cool will the spy be when he gets there?

33.A particle of mass min a rectangular box with dimensions x, y, and zhas ground state energy

where kis a physical constant. In Problem 29, Section 3, you were asked to min-imize the ground state energy subject to the fixed volume constraint V0 xyz

using substitution. Solve the same constrained optimization problem using the method of Lagrange multipliers.

E(x, y, z) k 2 8m

34.A rectangular building is to be constructed of material that costs $31 per square foot for the roof, $27 per square foot for the two sides and the back, and $55 per square foot for the fancy facing and glass used in constructing the front. If the building is to have a volume of 16,000 ft3, what dimensions will minimize the total cost of construction?

35.A storage shed is to be constructed of material that costs $15 per square foot for the roof, $12 per square foot for the two sides and back, and $20 per square foot for the front. What are the dimensions of the largest shed (in volume) that can be constructed for $8,000?

36.A manufacturer is planning to sell a new product at the price of $150 per unit and estimates that if x thousand dollars is spent on development and y thousand dollars is spent on promotion, approximately units of the product will be sold. The cost of manufacturing the product is $50 per unit. If the manufacturer has a total of $8,000 to spend on development and promotion, how should this money be allocated to generate the largest possible profit? [Hint:Profit

(number of units)(price per unit cost per unit) total amount spent on development and promotion.]

37.Suppose the manufacturer in Problem 36 decides to spend $8,100 instead of $8,000 on the development and promotion of the new product. Use the Lagrange multiplier

to estimate how this change will affect the maximum possible profit.

38. (a)If unlimited funds are available, how much should the manufacturer in Problem 36 spend on development and how much on promotion in order to generate the largest possible profit? [Hint:Use the methods of Section 3.]

(b)What is the value of the Lagrange multiplier that corresponds to the opti-mal budget in part (a)? Explain your answer in light of the interpretation of

as .

(c) Your answer to part (b) should suggest another method for solving the prob-lem in part (a). Solve the probprob-lem using this new method.

39.A consumer has $280 to spend on two commodities, the first of which costs $2 per unit and the second $5 per unit. Suppose that the utility derived by the consumer from x units of the first commodity and y units of the second is U(x, y) 100x0.25y0.75.

(a)How many units of each commodity should the consumer buy to maximize utility?

(b)Compute the marginal utility of money and interpret the result in economic terms.

40.A consumer has kdollars to spend on two commodities, the first of which costs a

dollars per unit and the second bdollars per unit. Suppose that the utility derived by

UTILITY UTILITY dM dk ALLOCATION OF UNRESTRICTED FUNDS MARGINAL ANALYSIS 320y y2 160x x4 ALLOCATION OF FUNDS CONSTRUCTION CONSTRUCTION

the consumer from x units of the first commodity and y units of the second commodity is given by the Cobb-Douglas utility function U(x, y) xy, where 0 1 and 1. Show that utility will be maximized when and

.

41.In Problem 40, how does the maximum output change if k is increased by 1 dollar?

In Problems 42 through 44, let Q(x, y)be a production function, where x and y

represent units of labor and capital, respectively. If unit costs of labor and capital are given by p and q, respectively, then px qy represents the total cost of production. 42.Use Lagrange multipliers to show that subject to a fixed production level c, the total

cost is minimized when

and Q(x, y) c

provided Qxand Qyare not both 0 and p 0 and q 0. (This is often referred

to as the minimum cost problem, and its solution is called the least-cost com-bination of inputs.)

43.Show that the inputs xand ythat maximize the production level Q(x, y) subject to a fixed cost ksatisfy

with px qy k

(Assume that neither pnor qis 0.) This is called a fixed budget problem. 44.Show that subject to the fixed production level Axyk, with 1, the cost

function C(x, y) pxqyis minimized when

45.Use Lagrange multipliers to find the possible maximum or minimum points on that part of the surface z xyfor which y x5x2. Then use your cal-culator to sketch the curve y x5 x 2 and the level curves to the surface

f(x, y) x yand show that the points you have just found do not represent local maxima or minima. What do you conclude from this observation?

46.A study conducted at a waste disposal site reveals soil contamination over a region that may be described roughly as the interior of the ellipse

x2 4 y2 9 1 HAZARDOUS WASTE MANAGEMENT x k A

where xand yare in miles. The manager of the site plans to build a circular enclo-sure to contain all polluted territory.

(a)If the office at the site is at the point S(1, 1), what is the radius of the small-est circle centered at S that contains the entire contaminated region? [Hint:

The function f(x, y) (x 1)2 (y 1)2measures the square of the dis-tance from S(1, 1) to the point P(x, y). The required radius can be found by maximizing f(x, y) subject to a certain constraint.]

(b)Read an article on waste management, and write a paragraph on how man-agement decisions are made regarding landfills and other disposal sites.*

47.Let P(K, L) be a production function, where Kand Lrepresent the capital and labor required for a certain manufacturing procedure. Suppose we wish to maximize

P(K, L) subject to a cost constraint, C(K, L) A, for constant A. Use the method of Lagrange multipliers to show that optimal production is attained when

that is, when the ratio of marginal production from capital to the marginal cost of capital equals the ratio of marginal production of labor to the marginal cost of labor.

In Exercises 48 and 49, use the formula for implicitly differentiating the function y f(x) given by the equation F(x, y) k.

48.Let F(x, y) x22xy y2.

(a)If F(x, y) kfor constant k, use the method of implicit differentiation devel- oped in Chapter 2 to find .

(b)Find the partial derivatives Fxand Fy and verify that

49.If F(x, y) k, where F(x, y) xln (x y), find .dy dx y x xexy2 dy dx Fx Fy dy dx dy dx Fx Fy ∂C ∂L ∂C ∂K ∂P ∂L ∂P ∂K MARGINAL ANALYSIS

* An excellent case study may be found in M. D. LaGrega, P. L. Buckingham, and J. C. Evans,

In Problems 50 through 53, use the method of Lagrange multipliers to find the indicated maximum or minimum. You will need to use the graphing utility or the solve application on your calculator.

50.Maximize f(x, y) exy xln subject to x y 4.

51.Minimize f(x, y) ln (x 2y) subject to xy y 5.

52.Minimize f(x, y) subject to x 2y 7.

53.Maximize f(x, y) subject to x2 2y2 1.

In Chapters 5 and 6, you integrated a function of one variable f(x) by reversing the process of differentiation, and a similar procedure can be used to integrate a function of two variables f(x, y). However, since two variables are involved, we shall integrate

f(x, y) by holding one variable fixed and integrating with respect to the other.

For instance, to evaluate the partial integral you would integrate with respect to x, using the fundamental theorem of calculus with yheld constant:

Similarly, to evaluate , you integrate with respect to y, holding xconstant:

In general, partially integrating a function f(x, y) with respect to x results in a function of yalone, which can then be integrated as a function of a single variable,

thus producing what we call an iterated integral Similarly, the iterated integral is obtained by first integrating with respect to y,

holding xconstant, and then with respect to x. For instance,

Double

Integrals over

Rectangular

Regions

5