HW6_Sol - E160 Operations Research I Spring 2009 Homework...

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E160 Operations Research I Spring 2009 Homework #6 Solution Chapter 12 Section 12.7 2. f(x 1 , x 2 ) = [3 - 2x 1 -2x 2 ] (2.5, 1.5) = [-2, -3]. To find a new point solve max -(0.5 - 2t) 2 - (2.5 - 2t) - (1.5 - 3t) 2 = f(t) t 0 f'(t) = 4(0.5 - 2t) + 2 + 6(1.5 - 3t) = 0 if 13 - 26t = 0 or t = .50 New Point = (2.5, 1.5) + .5 (-2, -3) = (1.5,0) Since (1.5, 0) = [0 0] we conclude the algorithm . Review Problems 6. f(x 1 , x 2 ) =[exp(-x 1 -x 2 )(1 -x 1 -x 2 )-1 exp(-x 1 -x 2 )(1 -x 1 -x 2 )]. (0,1) = [-1 0]. Thus new point is (-t, 1) where t 0 maximizes f(t) = (1 - t)e t -1 + t. f'(t) = (1 - t)e t-1 - e t-1 + 1 = 0 for 1 = te t-1 or t = 1. thus new point is (-1,1) f(-1, 1) = [0 1] so new point is [-1 1+t] and we choose t 0 to maximize h(t) = te -t + 1, h'(t) = -te -t + e -t = 0 for t = 1. Thus new point is (-1, 2). Section 12.8 2. We wish to maximize L 2/3 K 1/3 subject to 2L + K = 10. It is easier to maximize ln L 2/3 K 1/3 = (2/3)ln L + (1/3)ln K (this is a concave function so we know that Lagrange
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HW6_Sol - E160 Operations Research I Spring 2009 Homework...

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