QUESTION IS IN THE DESCRIPTION

Question

A city's transportation system is in financial difficulty and needs to raise its commuter rail fares. The current round trip fare is $6.00 and has 15,000 riders daily with half of them parking in the station's parking lot, costing $3. For every 50-cent increase in the fare, the city estimates it will lose 500 riders. What should the rail fare be for the city to maximize its daily revenue?

Answer 1

Currently the system is making 6(15000) + 3(7500) = $112500 per day.

Revenue = TicketSales + Parking

Let x = the multiplier

TicketSales = (6+.5x)(15000-500x)

Parking = 3(15000-500x)/2

(6+x/2)(1500-500x) + (3/2)(15000-500x) = -250x² + 3750x + 112500

This is an inverted parabola. The vertex can be found by x = -b/(2a) = -3750/(-500) = 7.5

So the new price of the ticket = 6+7.5/2 = $9.75

The new Revenue = $126575 per day

Which is $14075 more than before