HW6Sol - Fall 2009 IEOR 160 Industrial Engineering...

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Fall 2009 IEOR 160 October 17, 2009 Page 1 of 2 HOMEWORK 6 SOLUTIONS Section 12.7 2. f ( x 1 ,x 2 ) = [3 - 2 x 1 , - 2 x 2 ]. Then f (2 . 5 , 1 . 5) = [ - 2 , - 3]. To find the new point we need to maximize g ( t ) = f (2 . 5 - 2 t, 1 . 5 - 3 t ) = - (0 . 5 - 2 t ) 2 - (2 . 5 - 2 t ) - (1 . 5 - 3 t ) 2 . Now g 0 ( t ) = 4(0 . 5 - 2 t ) + 2 + 6(1 . 5 - 3 t ) = 0 for t = 0 . 5. Hence, the new point is (1 . 5 , 0). Since f (1 . 5 , 0) = [ - 2 , - 3] = 0, we conclude the algorithm and the maximum is (1.5,0). Section 12 Review 6. f ( x 1 ,x 2 ) = [ e - ( x 1 + x 2 ) (1 - x 1 - x 2 ) - 1 ,e - ( x 1 + x 2 ) (1 - x 1 - x 2 )]. Then f (0 , 1) = [ - 1 , 0]. To find the new point we need to maximize g ( t ) = f ( - t, 1) = ( - t + 1) e - ( - t +1) + t ). Now g 0 ( t ) = - te t - 1 + 1 = 0 for t = 1. Hence, the new point is ( - 1 , 1). f ( - 1 , 1) = [0 , 1]. We maximize now h ( t ) = f ( - 1 , 1 + t ) = te - t + 1. h 0 ( t ) = (1 - t ) e - t = 0 for t = 1. Thus, the new
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HW6Sol - Fall 2009 IEOR 160 Industrial Engineering...

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