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'm not sure if it's correct but I calculated 43.3 to be the price increases would maximize the revenue. What I did was plugged the 43.3 into the equation (2+0.15x) which is the cost of each item and (4000-40x)but i'm not sure if 8.50 is right for the fare and 2264 people will ride that day. Can you just check if I've done it right?
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! 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
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