Answer:
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
