A university is trying to determine what price to charge for tickets to football games. At a price of ​$30 per​ ticket, attendance averages 40,000 people per game. Every decrease of ​$5 adds 10,000 people to the average number. Every person at the game spends an average of ​$5.00 on concessions. What price per ticket should be charged in order to maximize​ revenue? How many people will attend at that​ price?

To determine the price per ticket that will maximize revenue, we need to find the ticket price that corresponds to the highest average number of attendees per game.

Let's break it down step-by-step:

Step 1: Start with the given information
- Price per ticket: $30
- Average attendance at this price: 40,000

Step 2: Calculate the change in attendance for every decrease of $5 in ticket price
- For every $5 decrease in ticket price, attendance increases by 10,000 people.
- So, for a $25 ticket price, attendance will be 40,000 + 10,000 = 50,000 people.
- For a $20 ticket price, attendance will be 50,000 + 10,000 = 60,000 people.
- And so on.

Step 3: Calculate the revenue for each ticket price
- Revenue = (Ticket price * Attendance) + (Concession spending per person * Attendance)

Let's calculate the revenue for each ticket price:

For $30 ticket price:
Revenue = ($30 * 40,000) + ($5 * 40,000) = $1,200,000 + $200,000 = $1,400,000

For $25 ticket price:
Revenue = ($25 * 50,000) + ($5 * 50,000) = $1,250,000 + $250,000 = $1,500,000

For $20 ticket price:
Revenue = ($20 * 60,000) + ($5 * 60,000) = $1,200,000 + $300,000 = $1,500,000

For $15 ticket price:
Revenue = ($15 * 70,000) + ($5 * 70,000) = $1,050,000 + $350,000 = $1,400,000

For $10 ticket price:
Revenue = ($10 * 80,000) + ($5 * 80,000) = $800,000 + $400,000 = $1,200,000

As we can see, the ticket prices of $25 and $20 result in the highest revenue of $1,500,000.

Therefore, the price per ticket that should be charged in order to maximize revenue is $25, and the number of people attending at that price would be 50,000.

To find the price per ticket that will maximize revenue, we need to determine the relationship between attendance and ticket price, as well as the relationship between attendance and revenue.

Let's start by analyzing the relationship between attendance and ticket price. We are given that at a price of ​$30 per​ ticket, attendance averages 40,000 people per game. We are also told that every decrease of ​$5 in ticket price adds 10,000 people to the average attendance.

By using this information, we can set up a table:

| Ticket Price ($) | Attendance (People) |
|------------------|---------------------|
| 30 | 40,000 |
| 25 | 50,000 |
| 20 | 60,000 |
| 15 | 70,000 |
| 10 | 80,000 |
| 5 | 90,000 |

Now, let's analyze the relationship between attendance and revenue. We know that each person at the game spends an average of ​$5.00 on concessions. So, revenue from ticket sales is calculated by multiplying the number of attendees by the ticket price.

| Ticket Price ($) | Attendance (People) | Revenue ($) |
|------------------|---------------------|-------------|
| 30 | 40,000 | 1,200,000 |
| 25 | 50,000 | 1,250,000 |
| 20 | 60,000 | 1,200,000 |
| 15 | 70,000 | 1,050,000 |
| 10 | 80,000 | 800,000 |
| 5 | 90,000 | 450,000 |

To determine the ticket price that maximizes revenue, we need to find the maximum value in the Revenue ($) column. From the table, we see that the maximum revenue occurs at a ticket price of $25, where the revenue is $1,250,000.

Therefore, the university should charge $25 per ticket to maximize revenue. At that price, 50,000 people will attend the football game.

x = number of $5 discounts

ticket price: 30-5x
attendance: 40000 + 10000x

revenue: (40000 + 10000x)(5+(30-5x))
max revenue at x=7
Huh? Pay people $5 to attend?

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