Maximizing revenue. Edwards University wants to determine what price to charge for tickets to football games. At a price of 518 per ticket, attendance averages 40,000 people per game. Every decrease of 53 to the ticket price adds 10,000 people to the average attendance. Every person at a game spends an average of 54.50 on concessions. What price per ticket should be charged to maximize revenue?
How many people will attend at that price?

Price per ticket to maximize revenue = $337.8

Total number of people at that price = 74000 people

Step-by-step explanation:

Let x represent the number of people as a result of decrease of 53 to the ticket price. Therefore, the price of a single ticket is;

M(x) = 518 - 53x

Similarly, we can say number of visitors of the game is;

T(x) = 40000 + 10000x

We are told that every person at a game spends an average of 54.50 on concessions. Therefore,

Total revenue will be;

R(x) = (M(x) × T(x)) + 54.5(T(x))

R(x) = ((518 - 53x) × (40000 + 10000x)) + 54.5(40000 + 10000x)

R(x) = (20720000 - 2120000x + 5180000x - 530000x²) + 2180000 + 545000x

R(x) = 20720000 + 3060000x - 530000x² + 2180000 + 545000x

R(x) = -530000x² + 3605000x + 22900000

To find the value of x for which the price is maximum, let's find the derivative of R(x) and equate to zero

dR/dx = -1060000x + 3605000

At dR/dx = 0,we have;

-1060000x + 3605000 = 0

1060000x = 3605000

x = 3605000/1060000

x = $3.4

Therefore price per ticket for the revenue to be maximum is;

M(3.4) = 518 - 53(3.4) = $337.8

Total attendance at that price will be;

T(3.4) = 40000 + 10000(3.4)

T(3.4) = 40000 + 34000

T(3.4) = 74000

12m

step-by-step explanation:

square root means have of a number so have of 3 is 1.5 then multiply it by 8 and you get your answer!

it will cost dollar $540

the living room is 3 which means if we divide 325/3 then we get $108

therefore if 1part of the living room cost $108 and 3 costed $325

then total 5part of the room will cost 5*108 which gives $540