SOLUTION: A charter flight charges a fare of $300 per person plus $6 per person for each unsold seat on the plane. If the plane holds 100 passengers and if x represents the number of u
Algebra ->
Finance
-> SOLUTION: A charter flight charges a fare of $300 per person plus $6 per person for each unsold seat on the plane. If the plane holds 100 passengers and if x represents the number of u
Log On
Question 1154438: A charter flight charges a fare of $300 per person plus $6 per person for each unsold seat on the plane. If the plane holds 100 passengers and if x represents the number of unsold seats, find the following.
A. An expression for the total revenue received for the flight. (Hint: Multiply the number of people flying, 100-x, by the price per ticket.)
B. The number of unsold seats that will produce the maximum revenue.
C.The maximum revenue. Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! A charter flight charges a fare of $300 per person plus $6 per person for each unsold seat on the plane.
If the plane holds 100 passengers and if x represents the number of unsold seats, find the following.
:
A. An expression for the total revenue received for the flight. (Hint: Multiply the number of people flying, 100-x, by the price per ticket.
total cost = cost per seat * no. of seats sold
f(x) = (300+6x)*(100-x)
FOIL
f(x) = 30000 - 300x + 600x - 6x^2
f(x) = -6x^2 + 300x + 30000; is the revenue equation
:
B. The number of unsold seats that will produce the maximum revenue.
Max will be on the axis of symmetry; x = -b/(2a), a=-6; b=300
x =
x = 25 seats unsold
:
C.The maximum revenue.
no. of seats sold: 100-25 = 75
seat cost: 300 + 6(25) = $450
therefore
75 * 450 = $33,750 is max revenue
:
Note: you can also substitute 25 for x in the original equation to get the max revenue
