The ticket price that would maximize the total revenue would be $ 23.
Given that a football team charges $ 30 per ticket and averages 20,000 people per game, and each person spend an average of $ 8 on concessions, and for every drop of $ 1 in price, the attendance rises by 800 people, to determine what ticket price should the team charge to maximize total revenue, the following calculation must be performed:
- 20,000 x 30 + 20,000 x 8 = 760,000
- 24,000 x 25 + 24,000 x 8 = 792,000
- 28,000 x 20 + 28,000 x 8 = 784,000
- 26,000 x 22.5 + 26,000 x 8 = 793,000
- 27,200 x 21 + 27,200 x 8 = 788,000
- 26,400 x 22 + 26,400 x 8 = 792,000
- 25,600 x 23 + 25,600 x 8 = 793,600
- 24,800 x 24 + 24,600 x 8 = 792,000
Therefore, the ticket price that would maximize the total revenue would be $ 23.
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Answer:
x = 4
Step-by-step explanation:
In 
Hence, by basic proportionality theorem:
Set x as adult tickets.
Set y as children's tickets.
x + y = 15
30x + 20y = 270
Solve for x in the first equation.
x + y = 15
x = 15 - y
Plug this into the second equation.
30x + 20y = 270
30(15 - y) + 20y = 270
450 - 30y + 20y = 270
450 - 10y = 270
-10y = -180
y = 18
If there is 18 childrens tickets, there should be -3 adult tickets.
This is impossible, and this impossible answer occured because the question is written wrong.
There are a total of 15 tickets
The smallest costing ticket is the childrens ticket, which costs 20$.
If he only bought children tickets, this would be 20x15 which is 300$.
300$ is over 270$, which makes the question impossible.
The answer is B. 5.3 miles per day * 17 days = 90.1 total miles.
