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Let's break down the problem and solve it step by step.

### Given:
- The number of shirts sold per week is given by the equation:
  \[
  q = -40x + 600
  \]
  where $ q $ is the number of shirts sold, and $ x $ is the price per shirt in dollars.
- Fixed costs: You are charged $400 per week for using the facilities.
- Variable costs: Each shirt costs you $5 to produce.

### 1. **Weekly Cost Function (as a function of $ x $)**:
The total weekly cost includes:
- The fixed cost of $400 per week.
- The variable cost of producing the shirts, which is $5 per shirt. Since you sell $ q $ shirts per week, the cost of producing the shirts is $ 5q $.

Using $ q = -40x + 600 $, the weekly cost $ C(x) $ can be expressed as:
\[
C(x) = 400 + 5q = 400 + 5(-40x + 600)
\]
Let's simplify this:
\[
C(x) = 400 + 5(-40x + 600) = 400 - 200x + 3000
\]
\[
C(x) = 3400 - 200x
\]

### 2. **Weekly Revenue Function (as a function of $ x $)**:
Revenue is calculated by multiplying the number of shirts sold by the price per shirt, i.e. $ \text{Revenue} = q \times x $. Substituting $ q = -40x + 600 $, the revenue $ R(x) $ becomes:
\[
R(x) = x(-40x + 600)
\]
Simplifying:
\[
R(x) = -40x^2 + 600x
\]

### 3. **Weekly Profit Function (as a function of $ x $)**:
Profit is revenue minus costs, i.e. $ P(x) = R(x) - C(x) $:
\[
P(x) = (-40x^2 + 600x) - (3400 - 200x)
\]
Simplifying:
\[
P(x) = -40x^2 + 600x - 3400 + 200x
\]
\[
P(x) = -40x^2 + 800x - 3400
\]

### 4. **Maximizing the Profit**:
To find the price $ x $ that maximizes the profit, we need to find the critical points by taking the derivative of $ P(x) $ and setting it to zero:
\[
P'(x) = -80x + 800
\]
Setting $ P'(x) = 0 $:
\[
-80x + 800 = 0
\]
\[
80x = 800
\]
\[
x = 10
\]

So, the price per shirt that maximizes the profit is **$10**.

### 5. **Maximum Profit**:
To find the maximum possible weekly profit, substitute $ x = 10 $ into the profit function $ P(x) $:
\[
P(10) = -40(10)^2 + 800(10) - 3400
\]
\[
P(10) = -40(100) + 8000 - 3400
\]
\[
P(10) = -4000 + 8000 - 3400
\]
\[
P(10) = 600
\]

Thus, the **maximum possible weekly profit** is **$600**.

---

Would you like further details or have any questions? 

Here are 5 related questions to expand your understanding:
1. What happens to the weekly profit if the price is set lower than $10?
2. How does the cost function change if the production cost per shirt increases to $6?
3. What is the break-even price where profit is zero?
4. How would the problem change if the fixed cost increased to $500?
5. Can the profit function $ P(x) $ ever be positive if the price exceeds $20?

**Tip:** When optimizing, always check if you're maximizing or minimizing by analyzing the second derivative of the function.

The number of shirts you can sell in one week is given by q = -40x + 600, where q is the number of shirts you can sell if you charge $x per shirt. The Student Council charges $400 per week for the facility, and the T-shirts cost $5 each. Find the weekly cost as a function of x, find the weekly profit, and determine the unit price to maximize profit.

Learn how to find the weekly cost, revenue, and maximum profit for T-shirt sales using algebraic and quadratic functions. This problem includes detailed formulas for profit maximization, cost, and revenue, and is ideal for high school students learning about optimization and derivatives.

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