A baseball team plays in a stadium that holds 56000 spectators. With the ticket price at $8 the average attendance has been 23000. When the price dropped to $7, the average attendance rose to 28000. Assume that attendance is linearly related to ticket price.




What ticket price would maximize revenue? $
  


To solve this, we need to determine the ticket price that maximizes revenue. Revenue ($R$) is given by:

\[
R = \text{Price} \times \text{Attendance}
\]

### Step 1: Find the linear relationship between attendance and price.
Let the price be $p$ and the attendance be $A$. We know:

- At $p = 8$, $A = 23000$.
- At $p = 7$, $A = 28000$.

The slope of the linear relationship ($m$) is:

\[
m = \frac{\Delta A}{\Delta p} = \frac{28000 - 23000}{7 - 8} = \frac{5000}{-1} = -5000
\]

So, the equation relating attendance to price is:

\[
A = -5000p + b
\]

Substitute one point ($p = 8, A = 23000$) to find $b$:

\[
23000 = -5000(8) + b \implies 23000 = -40000 + b \implies b = 63000
\]

Thus, the linear equation for attendance is:

\[
A = -5000p + 63000
\]

### Step 2: Write the revenue equation.
Revenue is the product of price ($p$) and attendance ($A$):

\[
R(p) = p \cdot A = p \cdot (-5000p + 63000)
\]

\[
R(p) = -5000p^2 + 63000p
\]

### Step 3: Maximize revenue by finding the vertex of the parabola.
The revenue function is a quadratic equation, and the vertex of a parabola in the form $ax^2 + bx + c$ occurs at:

\[
p = -\frac{b}{2a}
\]

Here, $a = -5000$ and $b = 63000$. Substituting:

\[
p = -\frac{63000}{2(-5000)} = \frac{63000}{10000} = 6.3
\]

### Step 4: Verify the price and calculate revenue.
The ticket price that maximizes revenue is:

\[
\boxed{6.30}
\]

At $p = 6.30$, the attendance is:

\[
A = -5000(6.3) + 63000 = -31500 + 63000 = 31500
\]

The maximum revenue is:

\[
R = 6.30 \cdot 31500 = 198450
\]

Thus, the maximum revenue is $\$198,450$ when the ticket price is $\$6.30$.

---

### Questions to explore:
1. What is the attendance at the maximum revenue price?
2. How would the revenue change if the price was set to $7 instead?
3. What happens to revenue if prices are raised above $8?
4. Could a non-linear attendance model yield different results?
5. What is the elasticity of attendance with respect to ticket price in this scenario?

**Tip:** In optimization problems, always confirm the vertex calculation for quadratic functions to ensure accuracy.

A baseball team plays in a stadium that holds 56000 spectators. With the ticket price at $8 the average attendance has been 23000. When the price dropped to $7, the average attendance rose to 28000. Assume that attendance is linearly related to ticket price. What ticket price would maximize revenue?

Discover how to maximize revenue for a baseball stadium with a detailed solution. This problem involves linear relationships between ticket price and attendance, and quadratic equations to find the optimal price of $6.30 for maximum revenue of $198,450. Suitable for Grades 10-12 students.

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