Question

A university is trying to determine what price to charge for tickets to football games. At a price of ​$24 per​ ticket, attendance averages 40 comma 000 people per game. Every decrease of ​$3 adds 10 comma 000 people to the average number. Every person at the game spends an average of ​$6.00 on concessions. What price per ticket should be charged in order to maximize​ revenue? How many people will attend at that​ price?

Answer 1

Answer:  Price per ticket should be charged in order to maximize​ revenue is $15.

70000 people will attend at this price.

Explanation:

Let 'x' represent the decrease .

Using the given information,

Price per ticket = 24 - 3x

Average no. of people that watch the game = 40000 + 10000x

Additional money spent by every person = 6(40000 + 10000x)

Revenue [R(x)] = Price per ticket $\times$ Average no. of people that watch the game + Additional money spent

Revenue [R(x)] = (24 - 3x)$\times$(40000 + 10000x) + 6(40000 + 10000x)

On solving the above equation we get ,

Revenue [R(x)] = -30000$x^{2}$ + 180000x + 1200000

In order to find the critical point we'll differentiate the following with respect to x;

R'(x) = -60000x + 180000

∵ R'(x) = 0  

x = 3

Thus, the price per ticket that should be charged in order to maximize​ revenue is (24 - 3$\times$3 = 24 - 9 = $15)

People that will attend at this price = (40000 + 10000$\times$3) = 70000