Respuesta :
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
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
Answer:
Scenario II i.e. when the price of ticket is $[tex]$245[/tex] will generate the greatest profit for the airline
Step-by-step explanation:
Scenario I
[tex]5000[/tex] tickets are sold at $ [tex]250[/tex] per ticket
The total money earned by the flight agency is
Number of tickets x price of each tickets
[tex]= 5000 * 250\\[/tex]
[tex]= 1250000[/tex] dollars
Scenario II
The price of each ticket is reduced by $[tex]5[/tex]
The price of new ticket is
[tex]250 -5[/tex]
[tex]= 245[/tex] dollars
The new number of tickets sold after reducing the price is
[tex]5000+ 500\\[/tex]
[tex]5500[/tex]
The total money earned by the flight agency is
Number of tickets x price of each tickets
[tex]5500 * 245\\[/tex]
[tex]1347500[/tex] dollars
Scenario II i.e. when the price of ticket is $[tex]$245[/tex] will generate the greatest profit for the airline