Calculus 3 Section 14.8: Lagrange Multipliers

Question 1: Explain why Solution 1 fails to find the correct solution.

Hints: Read Example 2 and Solution 1 (p. 873 - 874). The general purpose of this example should be taken to be that Lagrange multipliers can save us from this issue, rather than the fact that this problem can be solved in a slightly different fashion.

Chapter 14 Partial Derivatives 14.8 Lagrange Multipliers 873 Solution The problem asks us to find the minimum value of the function [op - Va-0+ (-07+ ( -0) - Vety+2 subject to the constraint that 2xty - z - 5 =0. Since |OP| has a minimum value wherever the function has a minimum value, we may solve the problem by finding the minimum value of f(x, y, 2) subject to the constraint 2x + y - z - 5 = 0 (thus avoiding square roots). If we regard x and y as the independent variables in this equation and write z as z = 2x + y - 5, our problem reduces to one of finding the points (x, y) at which the function h(x, y) = f(x, y, 2x + y - 5) =x + y' + (2x + y - 5) has its minimum value or values. Since the domain of & is the entire xy-plane, the First Derivative Test of Section 14.7 tells us that any minima that h might have must occur at points where h, = 2x + 2(2x + y - 5)(2) = 0, h, = 2y + 2(2x + y - 5) = 0. This leads to 10x + 4y = 20, 4x + 4y = 10, which has the solution WIL We may apply a geometric argument together with the Second Derivative Test to show that these values minimize h. The z-coordinate of the corresponding point on the plane z = 2x + y - 5 is + - 5 = 5Therefore, the point we seek is Closest point: P The distance from P to the origin is 5/V6 = 2.04. Attempts to solve a constrained maximum or minimum problem by substitution, as we might call the method of Example 1, do not always go smoothly. 1. 0, 0) (1, 0, D) EXAMPLE 2 Find the points on the hyperbolic cylinder x - 2- - 1 = 0 that are closest to the origin. Solution 1 The cylinder is shown in Figure 14.52. We seek the points on the cylinder clos- est to the origin. These are the points whose coordinates minimize the value of the function f(r. . ) = xty'+: Square of the distance subject to the constraint that x - z' - 1 = 0. If we regard x and y as independent vari- ables in the constraint equation, then FIGURE 14.52 The hyperbolic cylinder x - - 1 =0 in Example 2.874 Chapter 14 Partial Derivatives The hyperbolic cylinder x' - ?' = 1 and the values of f(x. y. z) = x3 + y' + 2 on the cylinder are given by the function On this part. On this part. h(x,y)=>+y+ (x- 1) =27+y-1. x- -Va+1 To find the points on the cylinder whose coordinates minimize f. we look for the points in the ry-plane whose coordinates minimize h. The only extreme value of h occurs where h = 4x = 0 and My = 2y = 0. that is, at the point (0, 0). But there are no points on the cylinder where both x and y are zero. What went wrong? What happened was that the First Derivative Test found (as it should have) the point 1=-1 in the domain of h where h has a minimum value. We, on the other hand, want the points on the cylinder where h has a minimum value. Although the domain of h is the entire xy-plane, the domain from which we can select the first two coordinates of the points (x, y, z) on the FIGURE 14.53 The region in the cylinder is restricted to the "shadow" of the cylinder on the xy-plane; it does not include the x)-plane from which the first two band between the lines x = -1 and x = 1 (Figure 14.53). coordinates of the points (r, y, z) on the We can avoid this problem if we treat y and z as independent variables (instead of x hyperbolic cylinder x- - 2' = 1 are and y) and express x in terms of y and z as selected excludes the band -1