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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?

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

70000 people will attend at this price.

Step-by-step 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
* Average no. of people that watch the game + Additional money spent

Revenue [R(x)] = (24 - 3x)
*(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
*3 = 24 - 9 = $15)

People that will attend at this price = (40000 + 10000
*3) = 70000

User Hari Menon
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