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Solution to Problem Set 09-10

# Solution to Problem Set 09-10 - Prof Gustavo Indart...

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Prof. Gustavo Indart Department of Economics University of Toronto ECO 100Y INTRODUCTION TO ECONOMICS Solutions to Problem Set 9-10 1. a) b) See diagram. i) Qo = 58 ii) Po = \$10.20 iii) TR = P * Q = 58 * \$10.20 = \$591.60 iv) TC = AC * Q = \$9.20 * 58 = \$533.60 v) Profits = TR - TC = 58*(\$10.20 - \$9.20) = \$58 or (Po - ACo)Qo. c) i) Q = 80 (at MR = 0) ii) Q = 88 (at P = SAC) d) i) Q D = Q S => P = \$7.50, Q = 84 ii) Profit = (P - AC) Q = approx (\$7.50 - \$7.10) 84 = \$33.60 iii) See shaded area on graph e) MC and AC each shift up by \$1.00 for each unit of output to SMC 1 and SAC 1 . MR = MC now at approximately Q = 55 giving P = \$10.50 and AC = \$10.50 Economic profits fall to perhaps 0. f) License fee is a fixed cost => no change in MC => non-parallel increase in AC with decreasing increase in AC with increased Q as the change in FC is spread over greater Q. MC = MR does not change => equilibrium is still at Q = 58 and P = \$10.20. AC has risen reducing economic profits by \$100 => Profit = \$58 - \$100 = -\$42 (negative economic profit). 0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 0 1 2 3 4 5 6 7 8 Pc 9 10 Po 11 12 1 13 14 15 MCo MC ACo 1 AC 1 P ACo D MR Q Qo Qc 1

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2 2. a), b) & c) MC AC D MR Po Qo Pm Pa Qm Qa Note that profit per unit = Po - ACo
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