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

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?

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

the answer is b; false. 2 dimensional things are a lot more different from 3 dimensional things.

i hope this ! : d

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