When the admission price for a baseball game was $4 per ticket, 45,000 tickets were sold. When the price was raised to $5, only 40,000 tickets were sold. Assume that the demand function is linear and that the variable and fixed costs for the ball park owners are $0.10 and $95,000 respectively.

(a)

Find the profit *P* as a function of *x*, the number of tickets sold.

*P*(*x*) =

To find the profit $ P(x) $ as a function of $ x $, the number of tickets sold, we'll first determine the price per ticket as a function of $ x $, then calculate the revenue and costs, and finally determine the profit function.

### Step 1: Determine the price function $ p(x) $

Given the dataTo find the profit $ P(x) $ as a function of $ x $, the number of tickets sold, we'll first determine the price per ticket as a function of $ x $, then calculate the revenue and costs, and finally determine the profit function.

### Step 1: Determine the price function $ p(x) $

Given the 
- When $ p = 4 $, $ x = 45,000 $
- When $ p = 5 $, $ x = 40,000 $

Assuming a linear demand function $ p(x) = mx + b $, we can find the slope $ m $ using the two points:

\[
m = \frac{5 - 4}{40,000 - 45,000} = \frac{1}{-5,000} = -\frac{1}{5,000}
\]

Using the point-slope form of the line equation:

\[
p(x) - 4 = -\frac{1}{5,000}(x - 45,000)
\]

Simplifying, we find:

\[
p(x) = -\frac{1}{5,000}x + 13
\]

### Step 2: Calculate Revenue and Costs

**Revenue $ R(x) $** is the price per ticket multiplied by the number of tickets sold:

\[
R(x) = p(x) \times x = \left(-\frac{1}{5,000}x + 13\right) \times x = -\frac{1}{5,000}x^2 + 13x
\]

**Total Costs $ C(x) $** consist of variable costs and fixed costs:

\[
C(x) = 0.10x + 95,000
\]

### Step 3: Determine the Profit Function $ P(x) $

Profit is given by the difference between revenue and total costs:

\[
P(x) = R(x) - C(x) = \left(-\frac{1}{5,000}x^2 + 13x\right) - (0.10x + 95,000)
\]

Simplifying the profit function:

\[
P(x) = -\frac{1}{5,000}x^2 + 13x - 0.10x - 95,000
\]

\[
P(x) = -\frac{1}{5,000}x^2 + 12.90x - 95,000
\]

### Final Answer:

\[
\boxed{P(x) = -\frac{1}{5,000}x^2 + 12.90x - 95,000}
\]

Would you like more details or have any questions?

### Related Questions:
1. What is the maximum profit, and how many tickets need to be sold to achieve it?
2. How does the break-even point change with different fixed costs?
3. What would the profit function look like if the variable costs increased to $0.20 per ticket?
4. How would the demand function change if the ticket sales were 50,000 at $6 per ticket?
5. What is the impact on profit if a discount of $1 per ticket is applied for every 10,000 tickets sold?

**Tip:** When dealing with profit functions, always check for the vertex of the parabola to find the maximum profit point.

When the admission price for a baseball game was $4 per ticket, 45,000 tickets were sold. When the price was raised to $5, only 40,000 tickets were sold. Assume that the demand function is linear and that the variable and fixed costs for the ball park owners are $0.10 and $95,000 respectively. (a) Find the profit P(x) as a function of x, the number of tickets sold.

Solve for the profit function P(x) based on a linear demand function for a baseball game. This detailed solution includes deriving the price function, calculating revenue, and determining the total profit with fixed and variable costs. Perfect for students in Grades 11-12.

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