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Unformatted text preview: ESE 415 Optimization Solution to Assignment 1 1. Consider the function f ( x ) defined on R 2 by f ( x,y ) = x 3 + e 3 y 3 xe y . Show that f ( x ) has exactly one critical point and that this point is a local minimizer but not a global minimizer of f ( x ). Solution: To find the critical point, we solve f ( x,y ) = 0. f ( x,y ) = f x = 3 x 2 3 e y = 0 f y = 3 e 3 y 3 xe y = 0 For x 6 = 0 , equation (1) times x minus equation (2) results in x 3 = e 3 y , so that x = e y . Substituting for e y into equation (2) results in x 2 = x for x 6 = 0. It follows that ( x,y ) = (1 , 0) is a critical point of f . If x = 0, then e y = 0 has no solution in R 2 . Therefore ( x,y ) = (1 , 0) is the unique critical point of f . To classify the critical point, examine the Hessian: H ( x,y ) = 2 f ( x,y ) = 6 x 3 e y 3 e y 9 e 3 y 3 xe y 2 f (1 , 0) = 6 3 3 6 Observe that 2 f (1 , 0) is positive definite, hence ( x,y ) = (1 , 0) is a local minimizer. For x < 0, the first principal minor 1 = 6 x < 0, in which case 2 f ( x,y ) is not positive definite. Therefore ( x,y ) = (1 , 0) is not a global minimizer. 2. Eigenvalues and Eigenvectors play important roles in optimization. (a) Find eigenvalues and eigenvectors for the following matrices: i) 2 2 2 , ii) 20 5 5 1 , iii) 3 2 2 2 7 2 2 2 3 Solution: 1 i) det  I A  = 2 2 + 2 = 2 + 2 + 4 = 1 3 i from( I A ) v i = 0 1 + 3 i 2 2 1 + 3 i + 2 x 1 x 2 = (...
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This document was uploaded on 09/11/2011.
 Spring '09

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