Question

An airline sells all tickets for a certain route at the same price. if it charges 250 dollars per ticket it sells 5000 tickets. for every 5 dollars the ticket price is reduced, an extra 500 tickets are sold. it costs the airline a hundred dollars to fly a person. what price will generate the greatest profit for the airline?

Answer 1

Scenario II i.e. when the price of ticket is $
`$245` will generate the greatest profit for the airline

Explanation:

Scenario I

`5000` tickets are sold at $
`250` per ticket

The total money earned by the flight agency is

Number of tickets x price of each tickets

`= 5000 * 250\\`

`= 1250000` dollars

Scenario II

The price of each ticket is reduced by $
`5`

The price of new ticket is

`250 -5`

`= 245` dollars

The new number of tickets sold after reducing the price is

`5000+ 500\\`

`5500`

The total money earned by the flight agency is

Number of tickets x price of each tickets

`5500 * 245\\`

`1347500` dollars

Scenario II i.e. when the price of ticket is $
`$245` will generate the greatest profit for the airline

Answer 2

let's look at the demand quantity as a function of price. Th problem tells us that q(250)= 500
the rate of change of q is a constant, thus, q is a line whose slope is -500/5=-100 tickets/ dollar.
therefore we shall have:
q(p)=5000-100(p-250)=3000-100p

the revenue for each price will be:
r(p)=p×q(p)=p(30000-100p)

next we get the total cost which is:
100×q, where q is the number of people who will fly.
but q=30000-100p
hence
c(p)=100(30000-100p)

the profit will be:
Profit=revenue-cost
=r(p)-c(p)=p(30000-100p)-100(30000-100p)
this will be written in quadratic form as:
(p-100)(30000-p)
next we find p that maximizes the profit function
(p-100)(30000-p)=-p²+30100p-3000000
when you take the derivative of the above, the maximum point will be at p=200, this will give us the profit of:
p(200)=-200²+30100(200)-3000000=1000000