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Solutions for Homework 4

Chapter 5: Problem 4

a. TFC|Q=10 = $100 = $100

b. TVC|Q=10 = $20*10 + $15*102 + $10*103 = $11,700.

c. TC|Q=10 = 100 + 20(10) + 15(10)2 + 10(10)3 = $11,800.

d. AFC|Q=10 = $100/10=$10.

e. AVC|Q=10 = TVC|Q=10 / 10 = $11,700/10 = $1,170.

f. SAC|Q=10 = AFC|Q=10 + AVC|Q=10 = $1,180.

g. MC|Q=10 = $20 + $30*10 + $30*102 = $3,320.

Chapter 5: Problem 6

Q FC VC TC AFC AVC ATC MC
0 15,000 0 15,000 - - - -
100 15,000 15,000 30,000 150 150 300 150
200 15,000 25,000 40,000 75 125 200 100
300 15,000 37,500 52,500 50 125 175 125
400 15,000 75,000 90,000 37.5 187.5 225 375
500 15,000 147,500 162,500 30 295 325 725
600 15,000 225,000 240,000 25 375 400 775

Chapter 5: Problem 7

a. Economies of scope exist since f – aQ1Q2 = 90 – (-0.5) Q1Q2 > 0 and fixed cost f  0 always.

b. Cost complementarities exist since a  0.5  0 .

c. Since a  0.5  0 , MC1 will increase if the division that produces product 2 is sold.

000 400 1.3 1.00.000.000 3. $1.80 100 80 2.67 2. AVC = PL/APL = 30/15 = 2 c.00 100 100 2.000 0 1.000 0.67 1.300 5 23 1.400 1. The diminishing marginal returns start at min MC: dMC/dQ = – 12 + 6Q = 0 => Q=2 g.000 800 1.000 b.200 2.000 25 33.5 b.67 100 40 1.000 => AFC = TFC/Q = 1.500 45 37. Problem Set: Problem 18 Marginal Units of Units of Average Units of Product of Fixed Variable Product of TFC TVC TC AFC SAC MC Output Variable Input Input Variable Input Input 100 0 0 – – 1.5 1.800 0.200 0. $2.200 10 27.000 1. SAC = TC/Q = (2.10 0.000/20 + 15 – 6*20 + 202 = 395 f. $2.600 0.18 2.40 c.000 – – – 100 20 600 30 30 1. MC = PL/MPL = 30/15 = 2 .44 100 60 2. MC = dTC/dQ = dTVC/dQ = 15 – 12Q + 3Q2 = 15 – 12*20 + 3*202 = 975 e.45 1. The diminishing average returns start at min AVC: dAVC/dQ = – 6 + 2Q = 0 => Q=3 Problem Set: Problem 20 a. TFC = PK*K = $100*10 = $1.50 1.000 + 15Q – 6Q2 + Q3)/Q = 2.43 1.5 1. AVC = TVC/Q = (15Q – 6Q2 + Q3)/Q = 15 – 6Q + Q2 = 15 – 6*20 + 202 = 295 d. $0.600 2.20 0.00 Problem Set: Problem 19 a.000 1.33 0.000/400 = 2.30 4.000 2.

175 17.000 – – – – 1 10.000 1.130 2.000 2 10.000 2.000 1.500 850 3.000 11.400 13.175 1.000 11.000 9.250 1.000 7.000 5.333 852 4.000 880 2.000 0 10. and LRAC = LRMC. the technology exhibits Constant Returns to Scale. Since CA|Q=8 = 128 > CB|Q=8 = 114.000 4.000 10.556 3.240 7 10.607 1.Problem Set: Problem 21 Q TFC TVC TC AFC AVC SAC MC 0 10.800 11.667 940 2.380 1.535 8 10. plant B should be built.556 12.400 2.640 1.880 1.433 1.025 2. If LRAC is flat.000 6 10. b.000 1.458 1. .185 756 4 10. Marginal returns to the variable input (marginal productivity) starts to diminish beyond a certain level of employment.040 1.400 14.000 1.800 5. $ units LRAC = LRMC Q Problem Set: Problem 23 a.865 Problem Set: Problem 22 a.000 3.400 2.640 15.000 900 5.350 844 5 10.040 19.900 800 3 10.

5*222 = 366 > 320 If only plant B were used: CB = 50 + 222 = 534 > 320 Problem Set: Problem 24 L Q APL MP L SMC AVC AFC STC 0 0 – – – – 0 672 2 10 5.072 .00 40.33 11.272 10 60 6.0 100.872 8 56 7.072 4 28 7.5*142 + 50 + 82 = 320 If only plant A were used: CA = 80 + 2*22 + 0.00 33.5Q2  50 + Q2 = CB => 0  0.472 6 48 8.10 10.7 3.0 22.0 2.00 5.00 4.00 28.00 67.00 25.5Q2 – 2Q – 30 => Q  10 plant A more efficient. Minimum cost condition: C = CA + CB = 80 + 2*14 + 0.25 1.00 14.00 2. CA = 80 + 2Q + 0.0 20.0 50.57 24.0 1.57 12.22 28.00 9. b. Minimum cost condition MCA = 2 + QA = 2QB = MCB and planned production QA + QB = 22 result in a system of equations in two unknowns with simultaneous solution: QA = 14 and QB = 8.0 1.5 133.33 38.2 1.2 2.0 40.672 12 63 5.00 10.