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Unformatted text preview: Homework 3
Exercise 5
There are multiple ways to approach this problem. The easiest is to use the fact that the constraint is an equality and substitute in. So...
x2 + y 2 = 1 x2 = 1 − y 2 f (x, y ) = 1 − 2y 2 It should be apparent that this function is maximized when y = 0. As such there are only two values of x which satisfy the constraint, −1 and 1. The value of the function at it's maximum is 1. You could also approach this problem with the method of Lagrange:
L = x2 − y 2 + λ(1 − x2 − y 2 ) ∂L = 2x − 2λx = 0 ∂x ∂L = −2y − 2λy = 0 ∂y (1) (2) ∂L = 1 − x2 − y 2 = 0 (3) ∂λ The rst holds if either x = 0, or λ = 1. The second equation holds if either y = 0, or λ = −1. In a maximization problem set up like this, a Lagrange multiplier must always be positive. So we know y = 0. Equation (3) then gives us that x = 1, −1, and the maximum value of f (x, y ) subject to the constraint is 1. 1 ...
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This note was uploaded on 11/07/2010 for the course ECO 33358 taught by Professor Mathevet during the Fall '10 term at University of Texas at Austin.
 Fall '10
 Mathevet

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