Question

# The E-Street Band has market power. Consumers demand tickets to each (monthly) live performance according to...

The E-Street Band has market power. Consumers demand tickets to each (monthly) live performance according to P = 15-0.1 * Q (MR = 15-0.2Q). The band faces a constant Marginal Cost of MC = \$0.20. Their total costs are TC = 150 + 0.2Q

a. In this setting, how many tickets should they sell to each concert to maximize profits? At what price should those tickets be sold? What are the band's profits? Graph this scenario.

b. What would be the band's profits if they could use 1^st Degree (Perfect) price discrimination? Graph this scenario.

c. Suppose that since the band has twice monthly performances (24 per year), they choose to use block/quantity pricing to sell "season" ticket packages to individuals. A ticket can then be used for ANY of the 24 concerts in a year. Suppose individual consumers have an annual demand for tickets represented by P = 4-0.2*Q. How much should the band charge for a bundle of tickets and how many tickets should be in each bundle, again assuming that MC is constant at \$0.20? Graph this scenario.

d. The bar where they band plays sells beer. When the band is performing each consumer's demand for beer is P = 7-2B, where B represents the quantity of beer consumed. The bar's marginal cost is \$1. The bar can use two-part pricing by charging a "cover" charge. How large should the cover charge be? How much should the bar charge for each beer? Graph this scenario.

a) Profit function is π = (15- 0.1Q)Q – 150-0.2Q

FOC:

dπ/dQ = 15 -0.2Q -0.2 = 0

This implies 14.80/0.2 = Q

Q* = 74 and P* = 7.6. substitute the value of P and Q in the profit function to find the profit

Profit = 562.4-150-14.8

Profit = \$397.60

b) Under 1st degree price discrimination the firm will sell till P=MC. The producer will eat up all the consumer surplus.

15-0.1Q = 0.2

14.8/0.1 = Q . Q* = 148

Profit = (15-0.2)*148 = \$2190.40

• For solution to other parts please post as separate question. Supposed to solve only these many parts.

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