revenue cost and profit application

Question

A community which is located in a resort area is trying to decide on the parking

fee to charge at the town-owned beach. There are other beaches in the area, and

there is competition for bathers among the different beaches. The town has

determined the following function which expresses the average number of cars

per day q as the function of parking fee p stated in cents.

q = 65000 – 13p

a. Determine the fee which should be charged to maximize daily beach

    revenues.

b. What is the maximum daily beach revenue expected to be.

Answer 1

$25 fee maximizes revenue

q = 65,000 - 13(2500) = 32,500

Revenue = pq = 25(32,500) = $812,500

Plot the equation q = 65,000 -13p

to maximize revenue, you want the maximum area under that line

That's the midpoint of the line (q,p) = (32,500, 2500) 65000/13(2) = 2500= $25.00 fee

q=65,000-13p

R=pq = 65,000p-13p^2, take the derivative and set equal to zero

R' = 65000-26p -0, solve for p

26p = 65000

p =65000/26 = 2500 = $25

max revenue = 2500 x 32,500 = 81250000 cents = $812,500