algebra word problem!

Question

A movie theater estimates that for each $0.50 increase in ticket price, the number of tickets sold decreases by 60. The current ticket price of $12.50 yields 1800 tickets sold. Set up and equation to represent the daily revenue. What price should the theater charge for tickets in order to maximize daily revenue?

Answer 1

Let x = number of $0.50 increases

 

R(x) = revenue = (number of tickets sold)(price/ticket)

 

                             = (1800 - 60x)(12.5 + 0.50x)

 

                             = -30x² + 150x + 22500

 

Maximum revenue when x = -150/[(2)(-30)] = -150/(-60) = 2.5

 

Price = 12.5 + 0.50(2.5) = $13.75

Answer 2

r = revenue

 

r = (12.50 + 0.5x)(1800 - 60x)

r = -30x² + 150x + 22500

r = -30(x² - 150x - 750)

r = -30(x - 30)(x + 25)

 

{-25, 30}

 

The zeros are -25 and 30.

The maximum occurs at 2.5.

Since the price increases by 50 cent increments, the maximum would happen at x = 2 or x = 3

The tickets can be either $13.50 or $14.00.