Question 871669
A bus company has 4000 passengers daily, each paying a fare of $2.
 For each $0.15 increase, the company estimates that it will lose 40 passengers per day. 
- What fare will maximize the revenue?
 How many passengers will ride at this price?
:
I think you have this right, using (4000-40x)(2-.15x), equation I got an axis
of symmetry of 43.3 also, A fare of $8.50, but I had 2268 passengers
A revenue of 8.5 * 2268 = $1978
:
But you can prove this to yourself, find the revenue when the fare is .15 less with 40 more passengers.
 8.35 * 2308 = $1971.80, slightly less
and
With a fare .15 more and 40 passengers less: 8.65 * 2228 = $1972.20, less also
You obviously have the maximum