Chapter 7
■
Section 4 Constrained Optimization: The Method of Lagrange Multipliers 551
LEVEL CURVES
45.
f ( x , y ) x
2
46.
f ( x , y ) 6 x
2
47.
f ( x , y ) 2 x
4 xy 7 y
2 x ln y
12 y
4 xy y
4 x 16 y 3 x
2
(11 y 18)
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 f at (0, 0).
y y f = 1 f =
1
2 x f =
1
3 f = –3 f = –2 f = –1 f = 3 f = 2
(b) f = 1 f = 3 f = 2 f = 1 x f = –3 f = –2 f = –1
(a)
4
In many applied problems, a function of two variables is to be optimized subject to a restriction or constraint on 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 x denotes the amount of money allocated to development, y the amount allocated 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 y 60,000.
For a geometric interpretation of the process of optimizing a function of two variables subject to a constraint, think of the function itself as a surface in three-dimensional space and of the constraint (which is an equation involving x and 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 minimum. The situation is illustrated in Figure 7.20.

552 Chapter 7 Calculus of Several Variables z
Unconstrained maximum
Constrained maximum y x
Constraint curve
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 solving the constraint equation for one of the variables and then substituting the resulting 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 technique called the method of Lagrange multipliers, in which the introduction of a third variable (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 relative extremum of the function f ( x , y ) subject to the constraint g ( x , y ) k must 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
F x f x g x
F y f y g y
F ( g k ) and solve the equations F x
0, F y
0, and
F
F x y f f x y g x g y
0
0
F ( g k ) 0
F or or or
0 simultaneously, as follows:
Finally, evaluate f ( a , b ) at each critical point ( a , b ) of F .
f x f y g x g y g k

Chapter 7
■
Section 4 Constrained Optimization: The Method of Lagrange Multipliers 553
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 f has a constrained maximum (minimum), it will be given by the largest (smallest) of the critical values f ( a , b ). Here is a summary of the procedure used in 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 f x
( x , y ) g x
( x , y ) f y
( x , y ) g y
( 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.
A geometric justification of the multiplier method is given at the end of this section. In the following example, the method is used to solve the problem from Example 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 fencing that will be needed to complete the job?
x y
Picnic area
Highway y
FIGURE 7.21
Rectangular picnic area.
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 ) x 2 y
The goal is to minimize f subject to the constraint that the area must be 5,000 square yards; that is, subject to the constraint xy 5,000

554 Chapter 7 Calculus of Several Variables
Let g ( x , y ) xy and use the partial derivatives f x
1 f y
2 g x y to obtain the three Lagrange equations
1 y 2 x
From the first and second equations you get and and xy g y
5,000 x
1 y and
(since y 0 and x 0), which implies that
2 x
1 y
2 x or x 2 y
Now substitute x 2 y into the third Lagrange equation to get
2 y
2
5,000 or y 50 and then use y 50 in the equation x 2 y to get x 100. It follows that x 100 and y 50 are the values that minimize the function f ( x , y ) x 2 y subject 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 x
2 y
2
8.
Solution
Let g ( x , y ) x
2 which implies that y
2 and use the partial derivatives f x y f y x to get the three Lagrange equations g x
2 x and g y
2 y y 2 x x 2 y and x
2 y
2
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
2 y x y x x y and or x
2
2 x y y
2

Chapter 7
■
Section 4 Constrained Optimization: The Method of Lagrange Multipliers 555
Now substitute x
2 y
2 into the third equation to get
2 x
2
8 or x 2
If x 2, it follows from the equation x
2 y
2 that y 2 or y 2. Similarly, if x 2, it follows that y 2 or y 2. Hence, the four points at which the constrained 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 x
2 y
2
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.
L
AGRANGE
M
ULTIPLIERS
FOR
F
UNCTIONS OF
T
HREE
V
ARIABLES
In each of the preceding examples, the first two Lagrange equations were used to eliminate the new variable , and then the resulting expression relating x and y was 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 problems 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 f x g x f x g y f z g z and g k
Here is an example of a problem involving this kind of constrained optimization.
x y z
A jewel box is to be constructed of material that costs $1 per square inch for the bottom, $2 per square inch for the sides, and $5 per square inch for the top. If the total volume is to be 96 in.
3 what dimensions will minimize the total cost of construction?
Solution
Let the box be x inches deep, y inches long, and z inches 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 1 yz 2(2 xy 2 xz ) 5 yz 6 yz 4 xy 4 xz abc
Bottom adddegddbgdddddec
Sides abc
Top

556 Chapter 7 Calculus of Several Variables
You wish to minimize C 6 yz 4 xy 4 xz subject to V xyz 96. The Lagrange equations are
C
C
C x y z
V x
V y
V z or or or
4 y 4 z ( yz )
6 z 4 x ( xz )
6 y 4 x ( xy ) and xyz 96. Solving each of the first three equations for , you get
4 y 4 z yz
6 z 4 x xz
6 y 4 x xy
By multiplying across each equation, you obtain
4
4 xyz xy
2
4 xz
2
6 yz
2
4 xyz
4 xyz 6 y
2 z 4 xyz
6 xyz 4 x
2 y 6 xyz 4 x
2 z which can be further simplified by first subtracting the common xyz terms from both sides of each equation
4 xz
2
4 xy
2
6 yz
2
6 y
2 z
4 x
2 y 4 x
2 z and then dividing z
2 from the third to get from both sides of the first equation, y
2 from the second, and x
2
4 x 6 y
4 x 6 z
4 y 4 z so that y z
2
3 x
Finally, by substituting these expressions into the constraint equation xyz 96, you find that x
2
3 x
2
3 x 96
4
9 x
3 x
3
96
216 so x 6 then y z
2
(6) 4
3
Thus, the minimal cost occurs when the jewel box is 6 inches deep with a square base, 4 inches on a side.

Chapter 7
■
Section 4 Constrained Optimization: The Method of Lagrange Multipliers 557
A
PPLICATIONS TO
E
CONOMICS
MAXIMIZATION OF UTILITY
In the next two examples, the method of Lagrange multipliers is used to solve constrained optimization problems from economics.
A utility function U ( x , y ) measures the total satisfaction or utility a consumer receives from having x units of one particular commodity and y units 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 x units of the first commodity and y units of the second commodity is given by the Cobb-Douglas utility function U ( x , y ) 10 x
0.6
y
0.4
. How many units of each commodity should the consumer buy to maximize utility?
Solution
The total cost of buying x units of the first commodity at $20 per unit and y units of the second at $30 per unit is 20 x 30 y . Since the consumer has only $600 to spend, the goal is to maximize utility U ( x , y ) subject to the budgetary constraint that 20 x
30 y 600.
The three Lagrange equations are
6 x
0.4
y
0.4
20 4 x
0.6
y
0.6
30 and 20 x 30 y 600
From the first two equations you get
9
6 y x
9 x
0.04
y
0.4
20
0.4
y
0.4
4 x
4 x
0.6
y
30
4 x
0.6
y
0.6
0.6
or y
4 x
9
Substituting this into the third equation, you get
20 x 30
4
9 x 600 from which it follows that x 18 and y
4
(18) 8
9
That is, to maximize utility, the consumer should buy 18 units of the first commodity and 8 units of the second.

558 Chapter 7 Calculus of Several Variables y
Optimal indifference curve: U ( x, y ) = C
20
(18, 8)
Budget line: 20 x + 30 y = 600 x
30
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 indifference curve U ( x , y ) C , where C U (18, 8) and the budgetary constraint 20 x 30 y
600, is sketched in Figure 7.22.
ALLOCATION OF RESOURCES 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 x thousand dollars is spent on development and y thousand on promotion, approximately f ( x , y ) 20 x
3/2 y 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 ) 20 x
3/2 y subject to the constraint g ( x , y ) 60, where g ( x , y ) x y . The corresponding Lagrange equations are
30 x
1/2 y 20 x
3/2 and x y 60

Chapter 7
■
Section 4 Constrained Optimization: The Method of Lagrange Multipliers 559
From the first two equations you get
30 x
1/2 y 20 x
3/2
Since the maximum value of f clearly does not occur when x 0, you may assume that x 0 and divide both sides of this equation by 30 x
1/2 to get y
2
3 x
Substituting this expression into the third equation, you get x
2
3 x 60 or
5
3 x 60 from which it follows that x 36 and y
2
(36) 24
3
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.
y
Optimal sales level: f ( x, y ) = 103,680
60
(36, 24)
Budget line: x + y = 60
60
FIGURE 7.23
Budgetary constraint and optimal sales level.
x

560 Chapter 7 Calculus of Several Variables
T
HE
S
IGNIFICANCE OF THE
L
AGRANGE
M
ULTIPLIER
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 M is the maximum (or minimum) value of f ( x , y ), subject to the constraint g ( x , y ) k . The Lagrange multiplier is the rate of change of M with respect to k . That is, dM dk
Hence, change in M resulting from a 1-unit increase in k
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
30 x
1/2 y 20 x
3/2 and x y 60 to find that the maximum value M of f ( x , y ) subject to the constraint x y 60 occurred when x 36 and y 24. To find , substitute these values of x and y into 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.
W HY THE M ETHOD
OF L AGRANGE
M ULTIPLIERS W ORKS
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 geometric 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
F x
F y

Chapter 7
■
Section 4 Constrained Optimization: The Method of Lagrange Multipliers 561
This result is true for any level curve of a function F whose partial derivatives exist
(provided F y
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 f that 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 ) k is equal to the slope of a level curve f ( x , y ) C .
y
Direction in which C increases
Point of tangency
Constraint curve: g ( x, y ) = k
Highest level curve, f ( x, y ) = C , intersecting the constraint x
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 g x g y f x f y or, equivalently, f x g x
If you let denote this common ratio, then f x g x and f y g y f y g y

562 Chapter 7 Calculus of Several Variables from which you get the first two Lagrange equations f x g x and f y g y
The third Lagrange equation g ( x , y ) k is simply a statement of the fact that the point of tangency actually lies on the constraint curve.
P . R . O . B . L . E . M . S 7.4
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 ) xy subject to the constraint x y 1.
2.
Find the maximum and minimum values of the function f ( x , y ) xy subject to the constraint x
2 y
2
1.
3.
Find the minimum value of the function f ( x , y ) x
2 xy 1.
y
2 subject to the constraint
4.
Find the minimum value of the function f ( x , y ) x
2 constraint 2 x y 22.
2 y
2 xy subject to the
5.
Find the minimum value of f ( x , y ) x
2 y
2 subject to the constraint x
2 y
2
4.
6.
Let f ( x , y ) 8 x
2
24 xy y
2
. Find the maximum and minimum values of the function f ( x , y ) subject to the constraint 8 x
2 y
2
1.
7.
Let f ( x , y ) x
2 y
2
2 y . Find the maximum and minimum values of the function f ( x , y ) subject to the constraint x
2 y
2
1.
8.
Find the maximum value of f ( x , y ) xy
2 subject to the constraint x y 1.
9.
Let f ( x , y ) 2 x
2
4 y
2
3 xy 2 x 23 y 3. Find the minimum value of the function f ( x , y ) subject to the constraint x y 15.
10.
Let f ( x , y ) 2 x
2 y
2
2 xy 4 x 2 y 7. Find the minimum value of the function f ( x , y ) subject to the constraint 4 x
2
4 xy 1.
11.
Find the maximum and minimum values of f ( x , y ) e xy subject to x
2 y
2
4.
12.
Find the maximum value of f ( x , y ) ln ( xy
2
) subject to 2 x
2
0 and y 0.
3 y
2
8 for x
13.
Find the maximum value of f ( x , y , z ) xyz subject to x 2 y 3 z 24.
14.
Find the maximum and minimum values of f ( x , y , z ) x 3 y z subject to z 2 x
2 y
2
.
15.
Find the maximum and minimum values of f ( x , y , z ) x 2 y 3 z subject to x
2 y
2 z
2
16.

Chapter 7
■
Section 4 Constrained Optimization: The Method of Lagrange Multipliers 563
CONSTRUCTION
CONSTRUCTION
POSTAL PACKAGING
16.
Find the minimum value of f ( x , y , z ) x
2 z
2
4.
y
2 z
2 subject to 4 x
2
2 y
2
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 rectangular parcel with two square sides that can be sent by fourth-class mail?
Girth = 4 x
Girth = 2
π x x y x y x
PROBLEM 19 PROBLEM 20
POSTAL PACKAGING
PACKAGING
PACKAGING
ALLOCATION OF FUNDS
20.
According to the postal regulation given in Problem 19, what is the largest volume of a cylindrical can that can be sent by fourth-class mail? (A cylinder of radius R and length H has volume R
2
H .)
21.
Use the fact that 12 fluid ounces is (approximately) 6.89
cubic 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 r and height h is r
2 h , that the circumference of a circle of radius r is 2 r , and that the area of a circle of radius r is r
2
.)
22.
A cylindrical can is to hold 4 cubic 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 x thousand dollars is spent on development and y thousand is spent on promotion, sales will be approximately f ( x , y ) 50 x
1/2 y
3/2 units. How much money should the manufacturer allocate to development and how much to promotion to maximize sales?

564 Chapter 7 Calculus of Several Variables
ALLOCATION OF FUNDS
MARGINAL ANALYSIS
SURFACE AREA OF THE
HUMAN BODY
24.
If x thousand dollars is spent on labor and y thousand dollars is spent on equipment, the output at a certain factory will be Q ( x , y ) 60 x
1/3 y
2/3 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 ) 0.0072
W
0.425
H
0.725
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 W H
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 L has volume V R
2
L and surface area S 2 RL 2 R
2
. The volume of a hemisphere of radius R is V
2
R
3
3 and its surface area is S 2 R
2
.
MICROBIOLOGY 27.
A bacterium is shaped like a cylindrical rod. If the volume of the bacterium is fixed, what relationship between the radius R and length H of the bacterium will result in minimum surface area?
Radius R
R
L
H
PROBLEM 27 PROBLEM 28
MICROBIOLOGY
OPTICS
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 L to achieve minimum surface area?
29.
The thin lens formula in optics says that the focal length L of a thin lens is related to the object distance d o and image distance d i by the equation
1 d o
1 d i
1
L
If L is fixed, what is the maximum distance s d o the image?
d i between the object and

Chapter 7
■
Section 4 Constrained Optimization: The Method of Lagrange Multipliers 565
CONSTRUCTION 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 cm
3
, 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?
Top view x y x x x
CONSTRUCTION
SPY STORY
PARTICLE PHYSICS
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 x
2 y
2
60 and that he is standing at the center (0, 0). He presses the stem on his special heat-detecting wristwatch and sees that the temperature at each point ( x , y ) in the room is given by
T ( x , y ) x
2 y
2
3 xy 5 x 15 y 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 m in a rectangular box with dimensions x , y , and z has ground state energy
E ( x , y , z ) k
2
8 m
1 x
2 y
1
2 z
1
2 where k is a physical constant. In Problem 29, Section 3, you were asked to minimize the ground state energy subject to the fixed volume constraint V
0 xyz using substitution. Solve the same constrained optimization problem using the method of Lagrange multipliers.

566 Chapter 7 Calculus of Several Variables
CONSTRUCTION
CONSTRUCTION
ALLOCATION OF FUNDS
MARGINAL ANALYSIS
ALLOCATION OF
UNRESTRICTED FUNDS
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 ft
3
, 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
320 y dollars is spent on promotion, approximately y 2 x
160 x
4 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 optimal budget in part (a)? Explain your answer in light of the interpretation of dM as .
dk
(c) Your answer to part (b) should suggest another method for solving the problem in part (a). Solve the problem using this new method.
UTILITY 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 )
100 x
0.25
y
0.75
.
(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.
UTILITY 40.
A consumer has k dollars to spend on two commodities, the first of which costs a dollars per unit and the second b dollars per unit. Suppose that the utility derived by

Chapter 7
■
Section 4 Constrained Optimization: The Method of Lagrange Multipliers 567
MINIMUM COST
FIXED BUDGET
MINIMUM COST
HAZARDOUS WASTE
MANAGEMENT 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 ) x y , where k
0 1 and 1. Show that utility will be maximized when x and a y k b
.
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
Q x p
Q y q and Q ( x , y ) c provided Q x and Q y are 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 combination of inputs.
)
43.
Show that the inputs x and y that maximize the production level Q ( x , y ) subject to a fixed cost k satisfy
Q x p
Q y q with px qy k
(Assume that neither p nor q is 0.) This is called a fixed budget problem.
44.
Show that subject to the fixed production level Ax y k , with 1, the cost function C ( x , y ) px qy is minimized when x k
A q p
, y k
A p q
45.
Use Lagrange multipliers to find the possible maximum or minimum points on that part of the surface z x y for which y x
5 x 2. Then use your calculator to sketch the curve y x
5 x 2 and the level curves to the surface f ( x , y ) x y and 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 x
2
4 y
9
2
1

568 Chapter 7 Calculus of Several Variables where x and y are in miles. The manager of the site plans to build a circular enclosure 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 smallest circle centered at S that contains the entire contaminated region? [ Hint:
The function f ( x , y ) ( x 1)
2
( y 1)
2 measures the square of the distance 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 management decisions are made regarding landfills and other disposal sites.*
MARGINAL ANALYSIS 47.
Let P ( K , L ) be a production function, where K and L represent 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
∂
P
∂
K
∂
C
∂
K
∂
P
∂
L
∂
C
∂
L 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 dy dx
F x
F y for implicitly differentiating the function y f(x) given by the equation F(x, y) k.
48.
Let F ( x , y ) x
2
2 xy y
2
.
(a) If F ( x , y ) k for constant k , use the method of implicit differentiation devel- dy oped in Chapter 2 to find .
dx
(b) Find the partial derivatives F x and F y and verify that dy dx
F x
F y
49.
If F ( x , y ) k , where F ( x , y ) xe xy
2 y x x ln ( x y dy
), find .
dx
* An excellent case study may be found in M. D. LaGrega, P. L. Buckingham, and J. C. Evans,
Hazardous Waste Management, McGraw-Hill, Inc., New York, 1994, pages 946
–
955.

Chapter 7
■
Section 5 Double Integrals over Rectangular Regions 569
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 ) e x y x ln y x subject to x y
51.
Minimize f ( x , y ) ln ( x 2 y ) subject to xy y 5.
52.
53.
Minimize
Maximize f f
(
( x x
,
, y y
)
)
1 x
2 xe x
2
3 xy y
1
2 subject to y subject to x
2
2 x y
2
2 y
1.
7.
4.
5
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
1
2 xy
2 dx you would integrate with respect to x , using the fundamental theorem of calculus with y held constant:
1
2 xy
2 dx
1
2 x
2 y
2 x 2 x 1
Similarly, to evaluate
1
2
(2)
2 y
2
1
2
(1)
2 y
2
3
2 y
2
1
1 xy
2 dy , you integrate with respect to y , holding x constant: y 1 1
1 xy
2 dy x
1
3 y
3 y 1 x
1
3
(1)
3 x
1
3
( 1)
3
2
3 x
In general, partially integrating a function f ( x , y ) with respect to x results in a function of y alone, which can then be integrated as a function of a single variable, thus producing what we call an iterated integral iterated integral f ( x , y ) dx dy .
Similarly, the f ( x , y ) dy dx is obtained by first integrating with respect to y , holding x constant, and then with respect to x . For instance,