It costs a bus company $225 to run a minibus on a ski trip, plus $30 per passenger. The bus has seating for 22 passengers, and the company charges $60 per fare if the bus is full. For each empty seat, the company has to increase the ticket price by $5. How many empty seats should the bus run with to maximize profit from this trip?
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Suppose the number of empty seats on the bus on a given trip is n.
Since there are 22 seats total, this means there are 22 - n seats occupied, that is, there are 22 - n passengers.
Then, the total cost for the bus company to run this trip, which is $225 plus $30 per passenger, is going to be $225 + $30(22 - n).
The company is going to charge each passenger $60 plus $5 for each empty seat. If there are n empty seats, each ticket's price is going to be $60 + $5n. Since there are 22-n passengers, the company will make the total of $(22-n)(60 +5n).
The profit can be found as the difference between the money made and the money spent, so the profit as the function of the number of empty seats n will be
`P=(22-n)(60 + 5n) - (225+30(22-n))`
Simplify this expression by opening the parenthesis:
`P=1320-60n + 110n - 5n^2 - 225-660+30n`
Combining the like terms results in
This is a quadratic function which has a maximum at the vertex.
The vertex for the general quadratic function `y=ax^2+bx+c`
has x-coordinate `x=-b/(2a)` .``
So for the profit function, the maximum will occur at `n=-80/(2*(-5)) = 8`
The bus company will maximize its profit if it runs the bus with 8 empty seats.
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