Homework3_2011_sol

# Homework3_2011_sol - ESE 415 Optimization Assignment 3 Due...

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Unformatted text preview: ESE 415 Optimization Assignment 3 Due: March 2, 2011 1. In each of the following problems fully justify your answer using optimality conditions. (a) Show that f ( x,y ) = ( x 2 − 4) 2 + y 2 has two global minima and one stationary point, which is neither a local maximum nor a local minimum. Solution: At critical points, ∇ f ( x,y ) = [ 4 x ( x 2 − 4) 2 y ] = [ ] ⇒ (0 , 0) (2 , 0) ( − 2 , 0) H ( x,y ) = [ 12 x 2 − 16 0 2 ] H (0 , 0) = [ − 16 0 2 ] H (2 , 0) = [ 32 0 2 ] H ( − 2 , 0) = [ 32 0 2 ] H (0 , 0) ⇒ indefinite neither max or min H (2 , 0) > ⇒ minimum H ( − 2 , 0) > ⇒ minimum f (2 , 0) = f ( − 2 , 0) ⇒ both (2,0) and (-2,0) are global minimums (b) Find all local mimina of f ( x,y ) = 1 2 x 2 + x cos y . Solution: At critical points ∇ f ( x,y ) = [ x + cos ( y ) − xsin ( y ) ] = [ ] ⇒ (0 , π 2 + nπ )where n ∈ Z , ( − 1 ,nπ )where n even, (1 ,nπ ) where n odd H ( x,y ) = [ 1 − sin ( y ) − sin ( y ) − xcos ( y ) ] H (0 , π 2 + nπ ) = [ 1 ± 1 ± 1 ] ,n ∈ Z H ( − 1 ,nπ ) = [ 1 0 0 1 ] ,n even H (1 ,nπ ) = [ 1 0 0 1 ] ,n odd H (0 , π 2 + nπ ) ⇒ indefinite neither max or min H ( − 1 ,nπ ) > ⇒ local minimum H (1 ,nπ ) > ⇒ local minimum (c) Find all local minima and all local maxima of f ( x,y ) = sin x + sin y + sin( x + y ) within the set { ( x,y ) | < x < 2 π, < y < 2 π } . 1 Solution: At critical points ∇ f ( x,y ) = [ cos ( x ) + cos ( x + y ) cos ( y ) + cos ( x + y ) ] = [ ] ⇒ ( π,π ) , ( π 3 , π 3 ) , ( 5 π 3 , 5 π 3 ) H ( x,y ) = [ − sin ( x ) − sin ( x + y ) − sin ( x + y ) − sin ( x + y ) − sin ( y ) − sin ( x + y ) ] H ( π,π ) = [ 0 0 0 0 ] H ( π 3 , π 3 ) = [ − √ 3 − √ 3 2 − √ 3 2 − √ 3 ] H ( 5 π 3 , 5 π 3 ) = [ √ 3 √ 3 2 √ 3 2 √ 3 ] H ( π,π ) ⇒ neither max or min H ( π 3 , π 3 ) < ⇒ local maximum H ( 5 π 3 , 5 π 3 ) > ⇒ local minimum 2. Use optimality conditions to show that for all x > 0 we have 1 x + x ≥ 2 . Solution: Let f ( x ) = 1 x + x . Then ∇ f ( x ) = − 1 x 2 + 1 = 0, hence x 2 = 1 so that x = 1, because x > 0. Now H ( x ) = 2 x 3 so H (1) = 2 > 0, and hence the global minimum occurs at x = 1. Because f (1) = 2, then x > ,f ( x ) = 1 x + x ≥ 2....
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Homework3_2011_sol - ESE 415 Optimization Assignment 3 Due...

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