SOLUTION: Please help me with this question (its a max/min problem):
It costs a bus company $225 to run a minibus on a ski trip, plus $40 per passenger. The bus has seating for 22 passeng
Question 926947: Please help me with this question (its a max/min problem):
It costs a bus company $225 to run a minibus on a ski trip, plus $40 per passenger. The bus has seating for 22 passengers, and the company charges $60 per fair 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?
I know to find profit it is revenue minus the cost, and cost is 225 + 40x
and i think revenue is (22 + x)(60 + 5x)
so i subtracted those: [(22 + x)(60 + 5x)] - (225 + 40x)
but when i completed the square i got a negative number. please help
You can put this solution on YOUR website! It costs a bus company $225 to run a minibus on a ski trip, plus $40 per passenger.
The bus has seating for 22 passengers, and the company charges $60 per fair 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?
:
If x = no. of empty seats,then it should be:
revenue is (22 - x)(60 + 5x) = 1320 + 110x - 60x - 5x^2
then: f(x) = [-5x^2 + 50x + 1320] - (225 + 40x)
f(x) = -5x^2 + 50x + 1320 - 225 - 40x
f(x) = -5x^2 + 10x + 1095
The axis of symmetry will be the max x
x =
x = 1 empty seat would give max profit ($1100)
