Could someone help me with this max revenue question? I keep getting the wrong answer and I don’t know why

Question

A bus service carries 10,000 people daily between Cybervelle and Steeptown, and the company has space to serve up to 15,000 people per day. The cost to ride the bus is $20. Market research shows that if the fare is increased by $0.50, 200 fewer people will ride the bus. What fare should be charged to get maximum revenue, given that the bus company must have at least $130,000 in fares a day to cover operating costs.

answer: new fare is $22.50 and max revenue is $202,500

Answer 1

Revenue = fare x number of riders

New fare will be = 20 + 0.5x (current fare of $20 plus x increments of $0.05)

Number of riders will then be = 10,000 - 200x (current number subtract 200 for each increment of x)

New revenue (R) = (20 + 0.5x)(10,000 - 200x)

R = -100x² + 1,000x + 200,000

Differentiate to find x-coordinate of maximum:

^dR/_dx = -200x + 1,000

-200x + 1,000 = 0

-200x = -1,000

x = 5

This gives a new fare of 20 + 0.5(5) = $22.50

And a maximum revenue of -100(5)² + 1,000(5) + 200,000 = $202,500