Question

A charter flight charges a fare of $300 per person plus $15 per person for each unsold seat on the plane. The plane holds 100 passengers. Let x represent the number of unsold seats. Find an expression for the total revenue received for the flight R(x). R(x)= Choose the correct graph of the function, R(x), below. The number of unsold seats that will produce the maximum revenue is (Round to the nearest whole number as needed.) The maximum revenue is $. (Round to the nearest whole number as needed.)

Answer 1

Answer:

The maximum revenue is $54,000 for 40 unsold seat plane.

Step-by-step explanation:

We are given the following in the question:

A charter flight charges a fare of $300 per person plus $15 per person for each unsold seat on the plane. The plane holds 100 passengers. Let x represent the number of unsold seats.

Total number of seats occupied = 100 - x

Revenue is given by  

$R(x) = (300+15x)(100-x)\\R(x) = 30000 + 1200x -15x^2$

First, we differentiate R(x) with respect to x, to get,

$\dfrac{d(R(x))}{dx} = \dfrac{d(30000 + 1200x -15x^2)}{dx} = 1200 - 30x$

Equating the first derivative to zero, we get,

$\dfrac{d(R(x))}{dx} = 0\\\\1200 - 20x= 0\\x = 40$

Again differentiation R(x), with respect to x, we get,

$\dfrac{d^2(R(x))}{dx^2} = -30$

At x = 40

$\dfrac{d^2(R(x))}{dx^2} < 0$

Thus, by double differentiation test maxima occurs at x = 40 for R(x).

Maximum revenue:

$R(40) = (300+15(40))(100-40) = 54000$

Thus, maximum revenue is $54,000 for 40 unsold seat plane.

The attached image shows the graph for the total revenue received for the flight