1. Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Lagrange Multipliers OBJECTIVES  Find maximum and minimum values using Lagrange multipliers.  Solve applied problems involving Lagrange multipliers. 9.4

2. Slide 6.5 - 2 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.

3. Slide 6.5 - 3 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Method of Lagrange Multipliers (continued) 2. Find the first partial derivatives F x, F y, and F λ. 3. Solve the system F x = 0, F y = 0, and F λ = 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.

4. Slide 6.5 - 4 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example 1: Find the maximum value of A(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 by F(x, y, λ) = xy – λ·(x + y – 20).

5. Slide 6.5 - 5 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example 1 (continued): 2. We find the first partial derivatives: F x = y – λ F y = x – λ F λ = – ( x + y – 20) 3. We set each derivative equal to 0 and solve the resulting system:

6. Slide 6.5 - 6 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

7. Slide 6.5 - 7 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example 2: Find the maximum value of f (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 by F(x, y, λ) = 3xy – λ(2x + y – 8).

8. Slide 6.5 - 8 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example 2 (continued): 2. We find the first partial derivatives: F x = 3y – 2λ F y = 3x – λ F λ = – (2x + y – 8) 3. We set each derivative equal to 0 and solve the resulting system:

9. Slide 6.5 - 9 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example 2 (continued): Solving the second equation for λ, we get λ = 3x. Substituting this into the first equation gives

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

11. Slide 6.5 - 11 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 in 3. 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 by s(h, r) = 2πrh + 2πr 2 subject to the volume constraint πr 2 h = 21.66 or πr 2 h – 21.66 = 0.

12. Slide 6.5 - 12 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 + 2πr 2 – λ·(πr 2 h – 21.66). 2. We find the first partial derivatives: S h = 2πr – λπr 2 S r = 2πh – 4πr – 2λπrh S λ = – (πr 2 h – 21.66)

13. Slide 6.5 - 13 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.

14. Slide 6.5 - 14 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.

15. Slide 6.5 - 15 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

16. Slide 6.5 - 16 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