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HW7_Sol

HW7_Sol - E160 Operations Research I Spring 2009 Homework#7...

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E160 Operations Research I Spring 2009 Homework #7 Solution Chapter 12 Section 12.9 2. K-T conditions yield (1) 1 - 2x 1 λ 1 = 0 (x 1 constraint) (2) -1 -2x 2 λ 1 = 0 (x 2 constraint) (3) λ 1 (1 - x 1 2 - x 2 2 ) = 0 , λ 1 0 Trying λ 1 >0, we find from (1) and (2) that x 1 = 1/2λ 1 and __ x 2 = -1/2λ 1 . Then (3) yields 1/2λ 1 2 = 1 or λ 1 = \/2 /2. __ __ Finally we find that x 1 = \2/2 and x 2 = -\/2/2. Since the constraint is convex and the objective function is linear, we know that we have found an optimal solution to the problem. 5. Let L 1 =Labor purchased at \$15 per hour L 2 = Labor purchased at \$25 per hour. We wish to solve max z = 270(L 1 + L 2 ) 1/2 K 1/3 - 15L 1 - 25L 2 - 5K st L 1 160, All variables nonnegative Ignoring the non-negativity restrictions, the K-T conditions yield (1) 135(L 1 + L 2 ) -1/2 K 1/3 - 15 - λ 1 = 0 (L 1 constraint) (2) 135(L 1 + L 2 ) -1/2 K 1/3 - 25 = 0 (L 2 constraint) (3) 90(L 1 + L 2 ) 1/2 K -2/3 - 5 = 0 (K constraint) (4) λ 1 (160 - L 1 ) = 0, λ 1 0 We try λ 1 >0 (which by (4) yields L 1 = 160). From (2) (L 1 + L 2 ) -1/2 = 5K -1/3 /27 or (L 1 + L 2 ) 1/2 = 27K 1/3 /5. From (3) 90(27)K -1/3 /5 = 5 or K = 918,330. From (2) (L 1 + L 2 ) 1/2 = 27K 1/3 /5 = 524.88 so L

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HW7_Sol - E160 Operations Research I Spring 2009 Homework#7...

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