143k views
4 votes
A baseball team plays in a stadium that holds 60000 spectators. With the ticket price at $9 the average attendance has been 23000. When the price dropped to $7, the average attendance rose to 30000. Assume that attendance is linearly related to ticket price. What ticket price would maximize revenue?

1 Answer

5 votes

Answer:

Explanation:

We can start by assuming that the relationship between the ticket price and attendance is linear, so we can write the equation for the line that connects the two data points we have:

Point 1: (9, 23000)

Point 2: (7, 30000)

The slope of the line can be calculated as:

slope = (y2 - y1) / (x2 - x1)

slope = (30000 - 23000) / (7 - 9)

slope = 3500

So the equation for the line is:

y - y1 = m(x - x1)

y - 23000 = 3500(x - 9)

y = 3500x - 28700

Now we can use this equation to find the attendance for any ticket price. To maximize revenue, we need to find the ticket price that generates the highest revenue. Revenue is simply the product of attendance and ticket price:

R = P*A

R = P(3500P - 28700)

R = 3500P^2 - 28700P

To find the ticket price that maximizes revenue, we need to take the derivative of the revenue equation and set it equal to zero:

dR/dP = 7000P - 28700 = 0

7000P = 28700

P = 4.10

So the ticket price that would maximize revenue is $4.10. However, we need to make sure that this price is within a reasonable range, so we should check that the attendance at this price is between 23,000 and 30,000:

A = 3500(4.10) - 28700

A = 5730

Since 23,000 < 5,730 < 30,000, we can conclude that the ticket price that would maximize revenue is $4.10.