Question

A local club is arranging a charter flight to Hawaii. The cost of the trip is ​$586 each for 80 ​passengers, with a refund of​ $5 per passenger for each passenger in excess of 80.

(a) Find the number of passengers that will maximize the revenue received from the flight.
(b) Find the maximum revenue.

Answer 1

a) The number of passengers that will maximize the revenue received from the flight is 99.

b) The maximum revenue is $48,609.

Explanation:

We have to analyse two cases to build a piecewise function.

If there are 80 or less passengers, we have that:

The cost of the trip is $586 for each passenger. So

`R(n) = 586n`

If there are more than 80 passengers.

There is a refund of $5 per passenger for each passenger in excess of 80. So the cost for each passenger is

`R(n) = (586 - 5(n-80))n = -5n^(2) +400n + 586n = -5n^(2) + 986n`.

So we have the following piecewise function:

`R(n) = \left \{ {{586n}, n\leq 80 \atop {-5n^(2) + 986n}, n > 80} \right`

The maxium value of a quadratic function in the format of
`y(n) = an^(2) + bn + c` happens at:

`n_(v) = -(b)/(2a)`

The maximum value is:

`y(n_(v))`

So:

(a) Find the number of passengers that will maximize the revenue received from the flight.

We have to see if
`n_(v)` is higher than 80.

We have that, for
`n > 80`,
`R(n) = -5n^(2) + 986n`, so
`a = -5, b = 986`

The number of passengers that will maximize the revenue received from the flight is:

`n_(v) = -(b)/(2a) = -(986)/(2(-5)) = 98.6`

Rounding up, the number of passengers that will maximize the revenue received from the flight is 99.

(b) Find the maximum revenue.

This is
`R(99)`.

`R(n) = -5n^(2) + 986n`

`R(99) = -5*(99)^(2) + 986*(99) = 48609`

The maximum revenue is $48,609.