Key to problem set _4

# Key to problem set _4 - 2 = 128,000 TR = 140,000 π =...

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Key to problem set #4 17.2 Q= 500 -4P, so P = 125 -0.25 Q. Therefore, MR = 125 – 0.50 Q. MR(100)= 125 – (0.50)(100)= 75 17.5 Q = 5000 – 50 P, so P = 100 – 0.02 Q. Therefore MR = 100 – 0.04 Q MC = 40. MR = MC: 100 – 0.04 Q = 40; Q = 1500; P = 100 – 0.02(1500) =70. TC(1500) = 40,000 +(40)(1500) = 100,000 AC = 100,000/1500 = 66.67 < P, so don’t shut down. 17.6 Q = 16,000 – 200 P; P = 80 – 0.005 Q; MR = 80 – 0.01Q; MC = 20 + 0.02 Q; so MR= MC: 80 – 0.01Q = 20 + 0.02Q; or Q = 2000; P = 80- 0.005(2000)= 70. TC(2000) = 48000 +20(2000) + 0.01(2000)
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Unformatted text preview: 2 = 128,000; TR = 140,000; π = 12,000. So don’t shut down. 17.11 Q = 1000-50P, so P = 20 – 0.02Q; MR = 20 – 0.04Q. MC = 1; 19 = 0.04Q, so Q = 475 P = 20 –(0.02)(475)= 10.50. With tax, MC t = 2; 18 = 0.04Q; Q t = 450; P = 20 – (0.02)(450) = 11. So the pass-through rate is ½ (as it always is in the case of a linear demand curve and constant MC.) 17.15 Q s = -100 + 0.01W; so AE = 1000 +10L; ME = 1000 +20 L . W L 50 100 2000 1500 dwl=(1/2)(500)(50)=12,500 AE (supply) ME...
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