Question

On a certain​ route, an airline carries 9000 passengers per​ month, each paying ​$30 . A market survey indicates that for each​ $1 increase in the ticket​ price, the airline will lose 50 passengers. Find the ticket price that will maximize the​ airline's monthly revenue for the route. What is the maximum monthly​ revenue?

Answer 1

Let's start by finding the current monthly revenue of the airline for the given number of passengers and ticket price:

Monthly revenue = number of passengers x ticket price

Monthly revenue = 9000 x $30

Monthly revenue = $270,000

Now, let's find the number of passengers for each $1 increase in the ticket price:

Number of passengers lost = 50

So, for a ticket price increase of $x, the number of passengers will be:

Number of passengers = 9000 - 50x

The ticket price will then be $30 + $x, and the revenue will be:

Revenue = (9000 - 50x) x ($30 + $x)

Revenue = 270,000 - 1500x + 30x + x^2

Revenue = x^2 - 1470x + 270,000

To find the ticket price that will maximize the revenue, we need to find the vertex of the parabola given by the revenue equation. The x-coordinate of the vertex is:

x = -b/2a

Where a = 1, b = -1470, and c = 270,000.

x = -(-1470)/2(1)

x = 1470/2

x = 735

So, the ticket price that will maximize the revenue is $30 + $735 = $765. The maximum monthly revenue is then:

Revenue = (9000 - 50(735)) x ($30 + $735)

Revenue = 105 x $765

Revenue = $80,325

Therefore, the ticket price that will maximize the airline's monthly revenue for the route is $765, and the maximum monthly revenue is $80,325.