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

Answer:

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

Step-by-step 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