This question is incomplete, the complete question is;
A university is trying to determine what price to charge for tickets to football games. At a price of $24 per ticket, attendance averages 40,000 people per game. Every decrease of $4 adds 10,000 people to the average number. Every person at the game spends an average of $4 on concessions. What price per ticket should be charged in order to maximize revenue? How many people will attend at that price?
Answer:
a) price per ticket = $18
b) 55,000 will attend at that price
Step-by-step explanation:
Given that;
price of ticket = $24
so cost of ticket = 24 - x
every decrease of $4 adds 10,000 people
for $1 adds 2,500 people
Total people = 40000 + 2500x
Revenue function = people × cost + people × 4
so
R(x) = (40000 + 2500x)(24 - x) + (40000 + 2500x)4
R(x) = 960,000 - 40000x + 60000x - 2500x² + 160,000 + 10000x
R(x) = 1120000 + 30000x - 2500x²
R(x) = -2500x² + 30000x + 1120000
now for maximum revenue; R'(x) is to zero
so
R'(x) = d/dx( -2500x² + 30000x + 1120000 ) = 0
⇒ -5000x + 30000 + 0 = 0
⇒ -5000x + 30000 = 0
⇒ 5000x = 30000
x = 30000 / 5000
x = 6
R'(X) = -5000 < 0 ⇒ MAXIMUM
∴ Revenue is maximum at x = 6
so cost of ticket ;
price per ticket = (24 - x) = 24 - 6 = $18
so price per ticket = $18
Number of people = 40000 + 2500x
= 40000 + (2500 × 6)
= 40000 + 15,000
= 55,000
Therefore, 55,000 will attend at that price
where 
is the acceleration
