Advanced Math question.

Question

The Thunderbirds professional indoor soccer team has 900 season ticket holders. The management of the team wants to increase the current price of $400. A survey indicated that for every $20 increase in price, the team will lose 15 season ticket holders. What is the maximum revenue the team could receive and at what price would this occur?

Answer 1

current price = $400 with 900 season ticket holders

at $420, they lose 15 season ticket holders

two points on the graph are (400,900) and (420,885)

others are (0,1200) and (1600,0) at $1600 no one buys a ticket

at $0 price, "only" 1200 want the free tickets

graphically this is a straight line with slope = -20/15 =-4/3

y intercept = 1600, x intercept = 1200

maximum revenue is the midpoint on the line segment or (1200/2, 1600/2) = (600,800)

max revenue = 800(600) = $480,000

P= -4T/3 + 1600

Revnue =R = PT = -4T^2/3 + 1600T

take the derivative and set = 0

R' = -8T/3 +1600 = 0

T = 1600(3/8) = 200(3) = 600

max R = -4(600)^2/3 +1600(600)= $480,000 at Price = $800 with 600 tickets sold