A baseball team plays in a stadium that holds 65,000 spectators. With the ticket price at , the
average attendance at recent games has been 36,000 . A market survey indicates that for every dollar
the ticket price is lowered, attendance increases by 4000 .
7. Find the price that maximizes revenue from ticket sales.

Question

A baseball team plays in a stadium that holds 65,000 spectators. With the ticket price at

, the
average attendance at recent games has been 36,000 . A market survey indicates that for every dollar
the ticket price is lowered, attendance increases by 4000 .
7. Find the price that maximizes revenue from ticket sales.

UpStudy AI · Accepted Answer

To find the ticket price that maximizes revenue, we can set up a revenue function based on the given information.

Let $$ p $$ be the price of a ticket in dollars. The current ticket price is $$ p = 15 $$ and the current attendance is 36,000. According to the market survey, for every dollar decrease in ticket price, attendance increases by 4,000.

Let $$ x $$ be the number of dollars the ticket price is lowered. Then the new ticket price can be expressed as:

$$
p = 15 - x
$$

The new attendance, given the increase of 4,000 spectators for each dollar decrease in price, can be expressed as:

$$
\text{Attendance} = 36,000 + 4,000x
$$

The revenue $$ R $$ from ticket sales can be calculated as the product of the ticket price and the attendance:

$$
R = \text{Price} \times \text{Attendance} = (15 - x)(36,000 + 4,000x)
$$

Now, we can expand this revenue function:

$$
R = (15 - x)(36,000 + 4,000x)
$$
$$
R = 15 \times 36,000 + 15 \times 4,000x - x \times 36,000 - 4,000x^2
$$
$$
R = 540,000 + 60,000x - 36,000x - 4,000x^2
$$
$$
R = 540,000 + 24,000x - 4,000x^2
$$

Now, we have a quadratic revenue function:

$$
R = -4,000x^2 + 24,000x + 540,000
$$

This is a downward-opening parabola, and the maximum revenue occurs at the vertex. The $$ x $$-coordinate of the vertex for a quadratic function $$ ax^2 + bx + c $$ is given by:

$$
x = -\frac{b}{2a}
$$

In our case, $$ a = -4,000 $$ and $$ b = 24,000 $$:

$$
x = -\frac{24,000}{2 \times -4,000} = \frac{24,000}{8,000} = 3
$$

This means the optimal price decrease is $$ x = 3 $$. Therefore, the price that maximizes revenue is:

$$
p = 15 - x = 15 - 3 = 12
$$

Thus, the price that maximizes revenue from ticket sales is:

$$
\boxed{12}
$$
The price that maximizes revenue is $$ \$12 $$.

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