a movie theater estimates that for each $0.50 increase in ticket price, the number of tickets sold decreases by 40. the current ticket price of $12.50 yields 12

Question

Need to know the price that the theater should charge for tickets in order to maximize daily revenue

Answer 1

graph this on the x y coordinates with ticket price on the y axis and number of tickets sold on the x axis

.50/-40 is the slope of the line giving the relationship between price and tickets

=-1/80

P=(-1/80)T + c

12.50 = (-1/80)(40) + c

25 = -1 + 2c

2c = 26

c = 13

P = (-1/80)T + 13

12.50 = -40/80 + 13

12.50 - 13 = -1/2

-1/2 = -1/2

at a price of $13, tickets sold are zero, with revenue =0

at zero price, tickets sold are 1040 = 13(80) with revenue =0

so, somewhere between $13 and $0 is the revenue maximizing price

Revenue = R = PT = [(-1/80)T + 13](T) = -T^2/80 +13T

take the derivative and set = 0

-T/40 + 13 = 0

-T = 13(40) = 520

P=(-1/80)(520) + 13 = -6.5+13 = $6.50

the revenue maximizing price

which is "coincidentally exactly half way between the two prices that give zero revenue, $13 and $0