1. Functions of Several Variables 13 Copyright © Cengage Learning. All rights reserved.

2. Lagrange Multipliers Copyright © Cengage Learning. All rights reserved. 13.10

3. 3 Understand the Method of Lagrange Multipliers. Use Lagrange multipliers to solve constrained optimization problems. Use the Method of Lagrange Multipliers with two constraints. Objectives

4. 4 Lagrange Multipliers

5. 5 Many optimization problems have restrictions, or constraints, on the values that can be used to produce the optimal solution. Such constraints tend to complicate optimization problems because the optimal solution can occur at a boundary point of the domain. In this section, you will study an ingenious technique for solving such problems - Method of Lagrange Multipliers. To see how this technique works, suppose you want to find the rectangle of maximum area that can be inscribed in the ellipse given by

6. 6 Let (x, y) be the vertex of the rectangle in the first quadrant, as shown in Figure 13.78. Figure 13.78 Lagrange Multipliers

7. 7 Because the rectangle has sides of lengths 2x and 2y, its area is given by f(x, y) = 4xy. Objective function You want to find x and y such that f(x, y) is a maximum. Your choice of (x, y) is restricted to first-quadrant points that lie on the ellipse Constraint Lagrange Multipliers

8. 8 Now, consider the constraint equation to be a fixed level curve of The level curves of f represent a family of hyperbolas f(x, y) = 4xy = k. In this family, the level curves that meet the given constraint correspond to the hyperbolas that intersect the ellipse. Moreover, to maximize f(x, y), you want to find the hyperbola that just barely satisfies the constraint. The level curve that does this is the one that is tangent to the ellipse (just touches it), as shown in Figure 13.79.  At that point gradients to f and g are parallel:  f(x, y) ||  g(x, y) =>  f(x, y) =  g(x, y) scalar is called Lagrange multiplier Lagrange Multipliers

9. 9

10. 10 Lagrange Multipliers

11. 11 Constrained Optimization Problems

12. 12 Example 1 – Lagrange Multiplier with 1 Constraint Find the maximum value of f(x, y) = 4xy where x > 0 and y > 0, subject to the constraint (x 2 /3 2 ) + (y 2 /4 2 ) = 1. Solution: To begin, let By equating  f(x, y) = 4yi + 4xj and  g(x, y) = (2 x/9)i + ( y/8)j, you can obtain the following system of equations. Divide!!!

13. 13 Example 1 – Solution From the first equation, you obtain = 18y/x, and substitution into the second equation produces Substituting this value for x 2 into the third equation produces cont’d

14. 14 So, Because it is required that y > 0, choose the positive value and find that So, the maximum value of f is Example 1 – Solution cont’d Sometimes messier!!!!

15. 15 Note how level lines/surfaces touch.

16. 16 Return to my example from 13.8: A rectangular box without a lid is to be made from 27 m^2 of cardboard. Which dimensions would maximize volume? Note how level lines/surfaces touch.

17. 17

18. 18 Ex 2 (modified) Divide!!! Note how level lines/surfaces touch.

19. 19 Note how level lines/surfaces touch.

20. 20 My example Find min value of function f on plane g. Note how level lines/surfaces touch.

21. 21

22. 22

23. 23

24. 24

25. 25 The Method of Lagrange Multipliers with Two Constraints

26. 26

27. 27 Example 5 – Optimization with Two Constraints Let T(x, y, z) = 20 + 2x + 2y + z 2 represent the temperature at each point on the sphere x 2 + y 2 + z 2 = 11. Find the extreme temperatures on the curve formed by the intersection of the plane x + y + z = 3 and the sphere. Solution: The two constraints are g(x, y, z) = x 2 + y 2 + z 2 = 11 and h(x, y, z) = x + y + z = 3.

28. 28 Using and  h(x, y, z) =  i +  j +  k you can write the following system of equations. T(x, y, z) = 20 + 2x + 2y + z 2 g(x, y, z) = x 2 + y 2 + z 2 = 11 h(x, y, z) = x + y + z = 3.

29. 29 By subtracting the second equation from the first, you can obtain the following system. (x – y) = 0 2z(1 – ) –  = 0 x 2 + y 2 + z 2 = 11 x + y + z = 3 From the first equation, you can conclude that = 0 or x = y. If = 0 => from 2 = 2 y +  =>  = 2=> from 2z = 2 z +  => z = 1 => x+y = 2 => y = 2 – x => 2x 2 -4x + 4 = 11-1 =>x=-1,3=> (3, –1, 1) and (–1, 3, 1). At both of these points T = 25 Example 5 – Solution cont’d

30. 30 If ≠ 0, then x = y and you can show that the critical points occur when x = y = and z = Finally, to find the optimal solutions, compare the temperatures at the four critical points. So, T = 25 is the minimum temperature and is the maximum temperature on the curve. Example 5 – Solution cont’d

31. 31 Mathematica Implementation

32. 32