Question

On a certain​ route, an airline carries 5000 passengers per​ month, each paying ​$60. 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

Answer:

$320000

Step-by-step explanation:

Given that on a certain route, an airline carries 5000 passengers per​ month, each paying ​$60.

A market survey indicates that for each​ $1 increase in the ticket​ price, the airline will lose 50 passengers

Let 1 dollar be increased x times (say)

Original revenue = 60(5000) =300000

After increase of x dollars, revenue would be a function of x

R(x) = $R(x) =(60+x) (5000-50x)\\\\R(x) = 300000+2000x-50x^2$

Use derivative test to find maximum

$R(x) = 300000+2000x-50x^2\\R'(x) = 2000-100x\\R"(x) = -100$

So maximum when I derivative is 0

i.e. when x = 20

Maximum revenue

$R(20) = (60+20)(5000-1000)\\= 320000$