Lagrange - Math 21a Handout on Lagrange Multipliers Fall...

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Math 21a Handout on Lagrange Multipliers - Fall 2000 The principal purpose of this handout is to supply some additional examples of the Lagrange multiplier method for solving constrained equations for three unknowns. In this regard, remember that the basic problem is to find the maxima, minima and/or stationary points of some given function f ( x , y , z ) on R 3 subject to the constraint that g ( x , y , z ) = 0 where g is another function. Note that these constrained problems typically arise in one of two ways: First, they can arise when looking for the extreme points of a function in some region of space; for in this case a search must be made for extreme points both inside the region and on its boundary. In particular, the problem of finding the extreme points on the boundary can be phrased as a constrained extremal problem. For example, imagine that the region in question is the solid Earth, and the function in question gives the temperature at each point. Then, a search for extreme points must look for interior extreme points and also extreme points on the surface. Constrained extremal problems also arise when looking for the stationary points of a given function’s restriction to some given surface in R 3 . For example, take the surface in question to be the surface of the earth, and consider finding the extreme points of the function whose values measure a year’s average of sun light intensity at each point. Anyway, what follows are some sample constrained problems. Example 1: Model the surface of the Earth as the sphere where x 2 + y 2 + z 2 = 1. Then, suppose that the average temperature at each point on the surface is given by the values of the function T = x 2 + 2 xy + y 2 - 2 z 2 . Find the points with the maximum and also the minimum temperature. Here is the solution: The constraint function is g ( x , y , z ) x 2 + y 2 + z 2 - 1, and the extreme points occur where T = λ∇ g for some constant λ . A computation finds: g = (2 x , 2 y , 2 z ) T = (2 x + 2 y , 2 y + 2 x , - 4 z ) (1) Thus, the equation T = λ∇ g holds if and only if 2 x + 2 y = λ 2 x 2 y + 2 x = λ 2 y -4 z = λ 2 z (2) The last point in (2) can hold only if λ = -2 or z = 0. I’ll treat these two possibilities in turn. Consider the case λ = -2. Then, the first and second points in (2) read 2 x + 2 y = -4 x 2 y + 2 x = -4 y (3) The latter points can be rephrased to read 6 x = -2 y 6 y = -2 x (4) This first line says that x = - y /3, while the second says that x = -3 y . Clearly, these two assertions are compatible only if x = y = 0. Finally, this last forces z = ±1 since x 2 + y 2 + z 2 = 1. Thus, I have found two stationary points, (0, 0, 1) and (0, 0, -1).
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Lagrange - Math 21a Handout on Lagrange Multipliers Fall...

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