Slide 6.5 - 1 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Constrained Maximum and Minimum Values: Lagrange Multipliers

OBJECTIVES Find maximum and minimum values

using Lagrange multipliers. Solve applied problems involving

Lagrange multipliers.

6.5

Slide 6.5 - 3 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

The Method of Lagrange Multipliers To find a maximum or minimum value of a function f (x, y) subject to the constraint g(x, y) = 0:

1. Form a new function:F(x, y, λ) = f (x, y) – λg(x, y).

The variable λ (lambda) is called a Lagrange multiplier.

6.5 Constrained Maximum and

Minimum Values: Lagrange Multipliers

Slide 6.5 - 4 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

The Method of Lagrange Multipliers (continued)2. Find the first partial derivatives Fx, Fy, and Fλ.

3. Solve the systemFx = 0, Fy = 0, and Fλ = 0,

Let (a, b, λ) represent a solution of this system. We normally must determine whether (a, b, λ) yields a maximum or minimum of the function f. For the problems in this text, we will specify that a maximum or minimum exists.

The method of Lagrange multipliers can be extended to functions of three (or more) variables.

6.5 Constrained Maximum and Minimum Values: Lagrange Multipliers

Slide 6.5 - 5 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example 1: Find the maximum value ofA(x, y) = xy

subject to the constraint x + y = 20.

First note that x + y = 20 is equivalent to x + y – 20 = 0.

1. We form the new function, F, given byF(x, y, λ) = xy – λ·(x + y – 20).

6.5 Constrained Maximum and Minimum Values: Lagrange Multipliers

Slide 6.5 - 6 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example 1 (continued): 2. We find the first partial derivatives: Fx = y – λ

Fy = x – λ

Fλ = – ( x + y – 20)

3. We set each derivative equal to 0 and solve the resulting system:

6.5 Constrained Maximum and Minimum Values: Lagrange Multipliers

y 0

x 0

(x y 20) 0

Slide 6.5 - 7 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example 1 (concluded): From the first two equations, we can see that x = λ = y. Substituting x for y in the last equation, we get

Thus, y = x = 10. The maximum value of A subject to the constraint occurs at (10, 10) and is

6.5 Constrained Maximum and Minimum Values: Lagrange Multipliers

10

202

020

x

x

xx

100

1010)10,10(

A

Slide 6.5 - 8 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example 2: Find the maximum value off (x, y) = 3xy

subject to the constraint 2x + y = 8.

Note that we rewrite 2x + y = 8 as 2x + y – 8 = 0.

1. We form the new function, F, given byF(x, y, λ) = 3xy – λ(2x + y – 8).

6.5 Constrained Maximum and Minimum Values: Lagrange Multipliers

Slide 6.5 - 9 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example 2 (continued): 2. We find the first partial derivatives: Fx = 3y – 2λ

Fy = 3x – λ

Fλ = – (2x + y – 8)

3. We set each derivative equal to 0 and solve the resulting system:

0)82(

03

023

yx

x

y

6.5 Constrained Maximum and Minimum Values: Lagrange Multipliers

Slide 6.5 - 10 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

6.5 Constrained Maximum and Minimum Values: Lagrange Multipliers

Example 2 (continued): Solving the second equation for λ , we get

λ = 3x.

Substituting this into the first equation gives

.2

63

0323

xy

xy

xy

Slide 6.5 - 11 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

6.5 Constrained Maximum and Minimum Values: Lagrange Multipliers

Example 2 (concluded): Substituting y = 2x into the third equation, we get

Then y = 2·2 = 4, and the maximum value of f subject to the constraint occurs at (2, 4) and is

.2

84

0822

x

x

xx

.24

423)4,2(

f

Slide 6.5 - 12 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example 3: The standard beverage can has a volume of 12 fl. oz, or 21.66 in3. What dimensions yield the minimum surface area? Find the minimum surface area. (Assume the shape of the can is a right circular cylinder.)

We want to minimize the function s, given bys(h, r) = 2πrh + 2πr2

subject to the volume constraint πr2h = 21.66 or πr2h – 21.66 = 0.

6.5 Constrained Maximum and Minimum Values: Lagrange Multipliers

Slide 6.5 - 13 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example 3 (continued): 1. We form the new function, S, given by

S(h, r, λ) = 2πrh + 2πr2 – λ·(πr2h – 21.66).

2. We find the first partial derivatives: Sh = 2πr – λπr2

Sr = 2πh – 4πr – 2λπrh

Sλ = – (πr2h – 21.66)

6.5 Constrained Maximum and Minimum Values: Lagrange Multipliers

Slide 6.5 - 14 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example 3 (continued): 3. We set each derivative equal to 0 and solve the

resulting system:

Since π is a constant, solve the first equation for r.

6.5 Constrained Maximum and Minimum Values: Lagrange Multipliers

2r r2 0

2h 4r 2rh 0

(r2h 21.66) 0

2or 0

02or 0

0)2(

rr

rr

rr

Slide 6.5 - 15 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example 3 (continued): Since r = 0 cannot be a solution to the problem, we will continue by substituting 2/λ into the second equation.

6.5 Constrained Maximum and Minimum Values: Lagrange Multipliers

2h 4 2

2

2

h 0

2h 8

4h 0

8

2h

4

h

Slide 6.5 - 16 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example 3 (continued): Since r = 2/λ and h = 4/λ, it follows that h = 2r. Substituting 2r for h in the third equation yields

6.5 Constrained Maximum and Minimum Values: Lagrange Multipliers

r2 (2r) 21.66 0

2r3 21.66

r3 10.83

r 10.83

3 1.51 in

Slide 6.5 - 17 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example 3 (concluded): Thus, when r = 1.51 in., we have h = 3.02 in, and the surface area is then a minimum and is approximately

6.5 Constrained Maximum and Minimum Values: Lagrange Multipliers

2 (1.51)(3.02) 2 (1.51)2 42.98 in2