CH12RP - 1 SOLUTIONS TO CHAPTER 12 MP REVIEW PROBLEMS 1 f(x...

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SOLUTIONS TO CHAPTER 12 MP REVIEW PROBLEMS 1. f''(x) = e -x 0 for all x. Thus f(x) is a convex function. 2. Let x = location of store. We wish to choose x to minimize f(x) = (x - 3) 2 + (x - 4) 2 + (x - 5) 2 + (x - 6) 2 + (x - 17) 2 f'(x) = 2[ x - 3 + x -4 + x - 5 + x - 6 + x - 17] = 0 for 3 + 4 + 5 + 6 + 17 x = --------------------- = 7. now f''(x) = 10>0 5 so x = 7 is a local minimum. Also f''(x)>0 implies that f(x) is convex so x = 7 does indeed minimize f(x). In general the store should be located at the arithmetic mean of the location of the n customers (i.e. x = (x 1 + x 2 + . .. x n )/n). 3a. Let R = units of raw material purchased. We wish to solve max z = (49 - x 1 )x 1 + (30 - 2x 2 )x 2 - 5R st (1) x 1 2R (2) x 2 R, All variables 0 Since the objective function is concave and the constraints are linear the K-T conditions will yield an optimal solution. The K- T conditions are (3) 49 - 2x 1 - λ 1 = 0 (4) 30 - 4x 2 - λ 2 = 0 (5) -5 + 2 λ 1 + λ 2 = 0 (6) λ 1 (x 1 - 2R) = 0 (7) λ 2 (x 2 - R) = 0 x 1 , x 2 , λ 1 , λ 2 0 Let's try λ 1 >0 and λ 2 = 0. Then (6) yields x 1 = 2R. From (5) λ 1 = 2.5. Then (3) yields x 1 = 23.25 and (4) yields x 2 = 7.5. Since x 1 = 2R we find that R = 11.625. All K-T conditions and original constraints are satisfied so we have found an optimal solution. Since the last unit of raw material purchased must have generated $5 in extra revenue, an additional unit of raw material would be worth slightly less than $5. 3b. R = 11.625, x 1 = 23.25, x 2 = 7.5. 4. For any two points on the curve y = |x| the line joining the two points is never below the curve. Thus |x| is a convex function of x. 5. x 1 = 5 - .618(5) = 1.91, x 2 = 0 + .618(5) = 3.09 f(x 1 ) = 2.08>f(x 2 ) = -.28 so new interval of uncertainty is [0, 3.09). x 3 = 3.09 - .618(3.09) = 1.18, x 4 = 1.91 f(x 3 ) = 2.15 >f(x 4 ) = 2.08 so new interval of uncertainty is [0, 1.91). x 6 = 1.18, x 5 = 1.91 - (.618)(1.91) = .73. f(x 6 ) = 2.15>f(x 5 ) = 1.66 so new interval of uncertainty is (.73, 1.91]. x 7 = 1.18 and x 8 = .73 + (.618)1.18 = 1.46. 1
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f(x 8 ) = 2.25>f(x 7 ) = 2.15 so new interval of uncertainty is (1.18, 1.91]. Now x 9 = 1.46 and x 10 = 1.18 +.618(.73) = 1.63 and f(x 9 ) = 2.25>f(x 10 ) = 2.23 so new interval of uncertainty is [1.18, 1.63). This interval has width less than .50, so we are finished. (actual maximum occurs for x = 1.5) 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).
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CH12RP - 1 SOLUTIONS TO CHAPTER 12 MP REVIEW PROBLEMS 1 f(x...

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