Answer: The revenue-maximizing price is $10.
Explanation:
Given that,
Inverse demand function: P = [tex]20 - \frac{Q}{1,000}[/tex]
Where,
P - Price per ride
Q - Number of rides per day
Revenue(R) = P × Q
= [tex]20 - \frac{Q}{1,000}[/tex] × Q
= [tex]20Q - \frac{Q^{2} }{1,000}[/tex]
Differentiating 'R' with respect to Q for calculating Marginal revenue(MR):
MR = [tex]20 - \frac{Q}{500}[/tex]
Here, MC = 0
MR = MC
[tex]20 - \frac{Q}{500}[/tex] = 0
Therefore, Q = 10,000
P = [tex]20 - \frac{Q}{1,000}[/tex]
= [tex]20 - \frac{10,000}{1,000}[/tex]
= $10
Hence, the revenue-maximizing price is $10.
