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

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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