h3e5 solutions

# h3e5 solutions - Homework 3 Exercise 5 There are multiple...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

## 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.

Ask a homework question - tutors are online