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 $16.50 yields 1800 tickets sold. Set up an equation to represent the daily revenue. What price should the theatre charge for tickets in order to maximize daily revenue

Answer 1

Answer:

The daily revenue equation is  29700 - 90x - 30x²

The price that should be charged to maximize daily revenue is $15.75

Explanation:

Revenue is a function of price multiplied by the quantity. Thus, the equation for revenue can be written as,

Let x be the number of times there is an increase in price of $0.5.

The price function is = 16.5 + 0.5x

The demand function is = 1800 - 60x

Revenue = (16.5 + 0.5x) * (1800 - 60x)

Revenue = 29700 - 990x + 900x - 30x²

Revenue = 29700 - 90x - 30x²

To calculate the price that will yield maximum revenue, we need to take the derivative of this equation of revenue.

d/dx = 0 - 1 * 90x° - 2 * 30x

0 = -90 -60x

90 = -60x

90 / -60 = x

x = -1.5

The price needed to maximize revenue is,

p = 16.5 + 0.5 * (-1.5)

p = 16.5 - 0.75

p = 15.75

The demand at this price is = 1800 - 60 * (-1.5)

Demand = 1800 + 90 = 1890