Question

On a certain​ route, an airline carries 7000 passengers per​ month, each paying ​$30. A market survey indicates that for each​ $1 increase in the ticket​ price, the airline will lose 100 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:

The ticket price that maximizes revenue is $50.

The maximum monthly revenue is $250,000.

Step-by-step explanation:

We have to write a function that describes the revenue of the airline.

We know one point of this function: when the price is $30, the amount of passengers is 7000.

We also know that for an increase of $1 in the ticket price, the amount of passengers will decrease by 100.

Then, we can write the revenue as the multiplication of price and passengers:

$R=p\cdot N=(30+x)(7000-x)$

where x is the variation in the price of the ticket.

Then, if we derive R in function of x, and equal to 0, we will have the value of x that maximizes the revenue.

$R(x)=(30+x)(7000-100x)=30\cdot7000-30\cdot100x+7000x-100x^2\\\\R(x)=-100x^2+(7000-3000)x+210000\\\\R(x)=-100x^2+4000x+210000\\\\\\\dfrac{dR}{dx}=100(-2x)+4000=0\\\\\\200x=4000\\\\x=4000/200=20$

We know that the increment in price (from the $30 level) that maximizes the revenue is $20, so the price should be:

$p=30+x=30+20=50$

The maximum monthly revenue is:

$R(x)=(30+x)(7000-100x)\\\\R(20)=(30+20)(7000-100\cdot20)\\\\R(20)=50\cdot5000\\\\R(20)=250000$