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

Answer:

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

b) The maximum revenue is $48,609.

Step-by-step 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:

$n_{v} = -\frac{b}{2a}$

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:

$n_{v} = -\frac{b}{2a} = -\frac{986}{2(-5)} = 98.6$

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.