COMM 290 Study Guide - Midterm Guide: Horse Length, Shadow Price, Feasible Region

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Published on 1 Oct 2018
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COMM 290 Midterm Solutions (2013)
Problem 1
(a) Skipping because of diagram, malfunctioning scanner
(b) True
(c) Produce 192 small and 240 large racquets subtract equation (10S + 12L = 4800) from
(10S + 30.5L = 9240), calculate S and L using algebra
(d) 10*192 + 15*240 = $5520
(e) 192/(192 + 240) = 44.44%
(f) 0.7625*240 = 183 kg
(g) ALLOY and TIME
(h) (10S + 12 L <= 4,800)/60 0.167S + 0.2L <= 80 hours
(i) E
(j) Zero
(k) Allowable increase: infinity; allowable decrease: 380 -240 = 140
(l) Subtract (0.25S + 0.7625L = 231)*40 from (10S + 12L = 4,800 + 1), plug L and S into MAX
equation, minus 5520 equals $0.839
(m) Calculate intersection of MIN % SMALL and ALLOY, plug into TIME equation, 4800 4600 =
allowable decrease of 740
(n) The new iso-profit line will fall directly on the TIME constraint (same slope). There will be
multiple/infinite optima from the old optimal solution point to the intersection of the time
constraint with the # of small racquets axis.
(o) TIME numerical evidence: 3*0.839*60 = 151.02 and 10*6.486 = 64.86, explanation: 180
minutes would increase profit by $151.02 and is within the allowable increase. Even though
alloy has a larger shadow price, 10kg increases profit by only $64.86. The goal is to
maximize profit!
Problem 2
(a) If they only produce large racquets, there would be unused resources and therefore not at
the optimal solution where profit is maximized ugly twin problem. At least another
point within the feasible region would have equal profits.
(b) Determine c in (-10/12) <= (-10/c) <= (-0.25/0.7625), 12 <= c <= 30.5
(c) Each small racquet requires 0.25kg of alloy to produce
(d) E
(e) 0.7S 0.3L >= 0
Problem 3
(a) 500, 300, and 200 pounds respectively of Bean 1, 2, and 3
(b) $3.00*300 = $900
(c) 500/(500 + 300+ 200) = 50%
(d) 0.6*500 + 0.84*300 + 0.54*200 = $660
(e) 94,400 80,000 = 4,400
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