Question

Suppose that the inverse demand for San Francisco cable car rides is p=10-(Q/1000), where p is the price per ride and Q is the number of rides per day. Suppose the objective of San​ Francisco's Municipal Authority​ (the cable car​ operator) is to maximize its revenuesL.
What is the​ revenue-maximizing price?

Answer 1

Answer:

Explanation:

Revenue is given by the number of rides per day (Q) multiplied by the price per ride (p):

$r=Q*p=Q*(10-\frac{Q}{1000}) \\R=10Q-\frac{Q^2}{1000}$

The number of rides 'Q' for which the derivate of the revenue function is zero is the revenue-maximizing number of rides:

$R(Q)=10Q-\frac{Q^2}{1000}\\R'(Q) = 0 = 10-\frac{Q}{500}\\Q=5000\ rides$

The price per ride at an activity of 5000 rides per day is:

$p(5,000) = 10 - \frac{5,000}{1,000}\\p=\ 5$

Therefore, the revenue-maximizing price is $5