Demand and Revenue Question

Question

Consider the demand for tickets to see a specific hockey team play. The price of the ticket can be related to the quantity demanded (q) by the function: p=214−0.01q. When the arena is not close to full capacity the total cost can be expressed by the function: Cost=96q+5,000,000.

Find marginal revenue (MR) as a function of quantity demanded.

MR= 

Let p∗ and q∗ be the price and quantity demanded where profit is maximized.

p∗=    q∗= 

The hockey players union has negotiated a deal requiring the team owner to pay an extra $1,000,000 a year in salaries to the players. What should the new ticket price (p1) be to ensure that profit is maximized.

p1= 

Answer 1

Let p denote price (of tickets) and q quantity (of tickets). Let R denote revenues. Then

R(q) = q*p(q) = q(214- 0.01q) = 214q -0.01q^2.

It is well known (or it should be!) that the marginal revenue is found by taking the derivative of the revenue function with respect to quantity; i.e.,

MR(q) = dR(q)/dq = 214 -0.02q.

To find the optimal number of tickets to sell and the price, we need to equate MR and MC (marginal cost).

Marginal cost is found as

MC(q) = dC(q)/dq = 96.

Hence, if we solve MR(q) = MC(q), we find that q* = 5900. Then we can find p* using either the equation p(q), that is:

p* = 214 - 0.01q* = 214- 0.01*5900 = 155.

Adding the $1,000,000 for players' salaries is not going to change the problem and its solution. Essentially, the only thing that changes are the fixed costs that go from $5,000,000 to $6,000,000 and that does not impact the marginal functions. Hence p(1) remains 155/ticket.