Question 1020083
A basketball team finds that if it charges $25 per ticket, the average attendance per game is 400.
 For each $0.50 decrease in price per ticket, attendance increases by 10.
 What ticket price yields the maximum revenue?
:
let x = the no. of .50 decreases and the no. of 10 people increases\
Revenue = price * no. of tickets
R(x) = (25-.5x)(400 + 10x)
FOIL
R(x) = 10000 + 250x - 200x - 5x^2
A quadratic equation
y = -5x^2 + 50x + 10000
Max revenue will be on the axis of symmetry; x = -b/(2a); a=-5; b=50
x = {{{(-50)/(2*-5)}}}
x = 5 decreases, 5*.50 = $2.50
Ticket price for max revenue: 25 - 25.0 = $22.50
:
:
Max rev, 50 ticket increase: 22.50 * 450 = $10,125