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.

Relax

Respuesta :

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

[tex]R(n) = 586n[/tex]

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

[tex]R(n) = (586 - 5(n-80))n = -5n^{2} +400n + 586n = -5n^{2} + 986n[/tex].

So we have the following piecewise function:

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

The maxium value of a quadratic function in the format of [tex]y(n) = an^{2} + bn + c[/tex] happens at:

[tex]n_{v} = -\frac{b}{2a}[/tex]

The maximum value is:

[tex]y(n_{v})[/tex]

So:

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

We have to see if [tex]n_{v}[/tex] is higher than 80.

We have that, for [tex]n > 80[/tex], [tex]R(n) = -5n^{2} + 986n[/tex], so [tex]a = -5, b = 986[/tex]

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

[tex]n_{v} = -\frac{b}{2a} = -\frac{986}{2(-5)} = 98.6[/tex]

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

(b) Find the maximum revenue.

This is [tex]R(99)[/tex].

[tex]R(n) = -5n^{2} + 986n[/tex]

[tex]R(99) = -5*(99)^{2} + 986*(99) = 48609[/tex]

The maximum revenue is $48,609.