Problems

Problems

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Chapter 11: Answers to Questions and Problems

1.
a. Since E = EF = EM, .
b. .
c. .

2.
a. P = \$60, Q = 4, and profits = 4(\$60 – \$20) = \$160.
b. Charge the maximum price on the demand curve starting at \$100 down to \$20 for each infinitesimal unit up to Q = 8 units. Profits are 8(\$100 – \$20)(.5) = \$320.
c. Charge a fixed fee of \$320 and a per-unit charge of \$20 per unit to earn total profits of \$320.
d. Create a package of 8 units and sell the package for \$480. Total profits are \$320.

3.
a. Second-degree price discrimination.
b. You will make three sales and earn \$8 in profit on the first sale (\$16 - \$8) and \$8 in profit on the last two sales (2(\$12 - \$8)) for a total profit of \$16.
c. Total profits under perfect price discrimination are \$8 + \$6 + \$4 + \$2 + \$0 = \$20, so this strategy would lead to an extra \$4. You could instead calculate the area of the triangle under demand and above the MC/AC curve to get 5(0.5(18-8)) = \$25. This formula would be correct if you could sell fractions of a unit, and generally will be the same as the discrete calculation for larger sales volumes.

4.
a. .
.
b. Here, there are two different groups with different (and identifiable) elasticities of demand. In addition, we must be able to prevent resale between the groups.

5.
a. Charge a per-unit fee equal of \$10, which equals marginal cost. At this price, you will sell 6 units. The fixed fee then should be (250 – 10)(6)(0.5) = \$720.
b. The optimal per-unit price is determined where MR = MC, or 250 - 80Q = 10. Solving yields Q = 3 units and P = \$130. The profits at this output and price are (\$130 - \$10)(3) = \$360. Thus, you earn \$360 more by two-part pricing.

6.
a. The inverse demand function is P = 160 – 2Q. Marginal cost is \$100. The optimal number of units in a package is that output where price equals marginal cost. Thus we set 160 – 2Q = 100 and solve to get the optimal number of units in a package, Q = 30 units.
b. The...