Please help me answer this calculus quadratics problem.

Question

A movie theater wants to determine the price per ticket that should be charged in order to maximize its revenue. The theater can accommodate 3000 movie goers per night, but the average attendance is about 1500 people per night. The average ticket price is about $8.00. Market research shows that, for each $0.50 reduction in the ticket price, attendance increases by 250 spectators. What should the theater charge per ticket to maximize the revenue ?

Answer 1

average attendance = 1500

capacity = 3000

average ticket price = $8

for each 50 cent reduction in price attendance increases by 250

What price maximizes revenue? $5.50

Revenue = Price x number of Tickets = R = PT

two points on the demand curve are (1500, 8) and (1000, 9) or using units of 1000

(1.5, 8) and (1,9)

as a 50 cent increase in price causes a 250 decrease in tickets sold

a $1 increase causes a 500 decrease in tickets

measure ticket sales in units of 1000

the slope = (9-8)/(1-1.5) = -1/.5 = -2

T = -2P + y where - 2 = slope and y = intercept

T = -2P + 19

when P=0, tickets at zero price only attract 19,000 customers

when P - $19, no one buys a ticket

halfway in between is (19/2)(11/2) = $52,250 max revenue

revenue maximizing price = $5.50

graph the demand curve. the midpoint gives the maximum revenue raising price and maximum revenue out put

Answer 2

Let charge per ticket is x. Then number of attendee is 1500 + 250(8 - x)/0.5 = 5500 - 500x.

So, revenue R(x) = (5500 - 500x)x = 5500x - 500x²;

R'(x) = 5500 - 1000x = 0; x = $5.50

R''(x) = - 1000 < 0. So, at x = $5.50 we have maximum revenue