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Unformatted text preview: 22M:270 Optimization Techniques Homework 1 answers 1. Find all the critical points of the following function f ( x , y ) = x 4 x 2 y + 2 y 2 2 y . Determine if they are local minima, local maxima, or saddle points, by looking at the Hessian matrices at the critical points. Evaluate the function at the critical points. Can you determine if the function is coercive (that is, f ( x , y ) → ∞ if p x 2 + y 2 → ∞ )? Do you think that you have found the global minimum? Explain why. First we look for stationary or critical points where ∇ f ( x ) = 0. Now ∂ f ∂ x ( x , y ) = 4 x 3 2 xy , ∂ f ∂ y ( x , y ) = x 2 + 4 y 2 , so the critical point occurs where x ( 4 x 2 2 y ) = = 4 y ( x 2 + 2 ) . Setting y = 1 4 ( x 2 + 2 ) and substituting into the first equation gives x ( 4 x 2 1 2 ( x 2 + 2 ))= 0. This gives either x = 0 or 7 2 x 2 1 = 0; substituting into the equation for y gives the following critical points: ( x , y ) = ( , 1 2 ) , (+ p 2 / 7 , 4 / 7 ) , or ( p 2 / 7 , 4 / 7 ) . Substitution back into the gradient formulas confirms that these are critical points. The Hessian matrix is ∇ 2 f ( x , y ) = ∂ 2 f / ∂ x 2 ∂ 2 f / ∂ x ∂ y ∂ 2 f / ∂ y ∂ x ∂ 2 f / ∂ y 2 , which is symmetric since ∂ 2 f / ∂ x ∂ y = ∂ 2 f / ∂ y ∂ x . That is, ∇ 2 f ( x , y ) = 12 x 2 2 y 2 x 2 x 4 ....
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This note was uploaded on 04/01/2012 for the course 22M 174 taught by Professor Davidstewart during the Spring '12 term at University of Iowa.
 Spring '12
 DavidStewart

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