A baseball team plays in a stadium that holds 52000 spectators. With the ticket price at the average attendance has been 23000 . When the price
dropped to , the average attendance rose to 26000 . Assume that attendance is linearly related to ticket price.
What ticket price would maximize revenue?

Ask by Page Pritchard. in the United States

Mar 23,2025

Question

A baseball team plays in a stadium that holds 52000 spectators. With the ticket price at the average attendance has been 23000 . When the price
dropped to , the average attendance rose to 26000 . Assume that attendance is linearly related to ticket price.
What ticket price would maximize revenue?

Ask by Page Pritchard. in the United States

Mar 23,2025

UpStudy AI · Accepted Answer

Let the ticket price be $$ x $$ (in dollars) and the attendance be $$ A(x) $$. We are given two data points:

- When $$ x = 8 $$, $$ A(8) = 23000 $$.
- When $$ x = 7 $$, $$ A(7) = 26000 $$.

Since attendance is linearly related to ticket price, we can write
$$
A(x) = ax + b.
$$

**Step 1. Determine the linear function parameters**

Using the two points, first find the slope $$ a $$:
$$
a = \frac{A(7) - A(8)}{7 - 8} = \frac{26000 - 23000}{7 - 8} = \frac{3000}{-1} = -3000.
$$

Now, substitute $$ a = -3000 $$ and one of the points into $$ A(x) = -3000x + b $$ to find $$ b $$. Using the point $$ (8, 23000) $$:
$$
23000 = -3000(8) + b.
$$
$$
23000 = -24000 + b \quad \Longrightarrow \quad b = 23000 + 24000 = 47000.
$$

Thus, the attendance function is:
$$
A(x) = -3000x + 47000.
$$

**Step 2. Express the revenue function**

Revenue $$ R(x) $$ is the product of the ticket price and the number of attendees:
$$
R(x) = x \times A(x) = x(-3000x + 47000) = -3000x^2 + 47000x.
$$

**Step 3. Find the ticket price that maximizes revenue**

The function $$ R(x) = -3000x^2 + 47000x $$ is a quadratic function that opens downward (since the coefficient of $$ x^2 $$ is negative). Its maximum occurs at the vertex. For a quadratic in the form $$ R(x) = Ax^2 + Bx + C $$, the vertex occurs at:
$$
x = -\frac{B}{2A}.
$$

Here, $$ A = -3000 $$ and $$ B = 47000 $$. Therefore,
$$
x = -\frac{47000}{2(-3000)} = \frac{47000}{6000} = \frac{47}{6}.
$$

In decimal form,
$$
\frac{47}{6} \approx 7.83.
$$

**Conclusion**

The ticket price that maximizes revenue is approximately $$ \$7.83 $$.
The ticket price that maximizes revenue is approximately \$7.83.

A baseball team plays in a stadium that holds 52000 spectators. With the ticket price at the average attendance has been 23000 . When the price
dropped to , the average attendance rose to 26000 . Assume that attendance is linearly related to ticket price.
What ticket price would maximize revenue?

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A baseball team plays in a stadium that holds 52000 spectators. With the ticket price at the average attendance has been 23000 . When the price dropped to , the average attendance rose to 26000 . Assume that attendance is linearly related to ticket price. What ticket price would maximize revenue?

Upstudy AI Solution

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A baseball team plays in a stadium that holds 52000 spectators. With the ticket price at the average attendance has been 23000 . When the price
dropped to , the average attendance rose to 26000 . Assume that attendance is linearly related to ticket price.
What ticket price would maximize revenue?