Chapter 11: Answers to Questions and Problems
a. Since E = EF = EM, .
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.
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.
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.
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.
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.