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

part a

x     y

0     8

1     7.5

2     7

3     6.5

4     6

5     5.5  

for every hour i deducted .5 inches from the length of the candle

part b

it is a function because for every x value there is only one y value

part   c

for every x value there is only one y value

x       y

0       8

1       7.6

2       7.2

3       6.8

step-by-step explanation: