Question

Concert ticket prices are $18 for a screening with 500 attendees. For each $1 increase in ticket price, attendance decreased by 5 people. What should the ticket price be to maximize revenue?

Answer 1

$59

Explanation:

Here you need to get the demand function to get the profit function and then maximize it.

For a price of $18 they sell 500 tickets. Then if the price rises up to $19 they sell 5 less tickets, it is, 495 tickets.

So, call D(p) the Demand function that depends on price p. The sentences above tell us:

D(18) = 500

D(19) = 495

We can think on a general demand function with the form:

D(p) = -ap + b

Where a and b are parameters we need to find, and we put a - sign before p to indicate that increases in price diminishes the demand. Replacing the previous values:

D(18) = - a(18) + b = 500 [1]

D(19) = - a(19) + b = 495 [2]

If we subtract [1] - [2]:

D(18) - D(19)= - a(18) + b - (- a(19) + b) = 500 -495

- 18a + b +19a - b = 5

a = 5

a = 5 (aprox)

Now we need b. We can get it by replacing tha a value in [1] or [2]:

-18(5) + b =500

90 + b = 500

b = 500 +90

b = 590

So, D(p) = -5p + 590

Now we need profit function. It is the sales (demand D) multiplied by price of sale p:

Prof = D(p) * p

Prof = [-5p + 590]*p

Prof (p) = -5p^2 + 590p

For maximize it just derivate and equal to 0:

Prof'(p) = 0

-10p + 590 = 0

590 = 10p

590 / 10 = p

59 = p

At price of $59 revenues are maximized