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

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

answer; (cash) is the best out of the above; good luck