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The problem in the image can be summarized as follows:

- The manufacturer produces souvenir T-shirts at a cost of $2 apiece.
- The current price of the T-shirts is $5, and at this price, 4000 shirts are sold each month.
- For each $1 increase in the price, 400 fewer shirts are expected to be sold each month.
- The task is to find the price at which the manufacturer should sell the T-shirts to maximize profit.

### Let's solve this step by step.

1. **Define Variables:**
   - Let $ p $ be the price of the T-shirts.
   - Let $ q(p) $ be the quantity of shirts sold at price $ p $.

2. **Revenue Function:**
   The revenue $ R(p) $ is the product of the price $ p $ and the quantity sold $ q(p) $.
   \[
   R(p) = p \cdot q(p)
   \]

3. **Quantity Function:**
   Initially, at a price of $5, the quantity sold is 4000. For each $1 increase in price, 400 fewer shirts are sold, so the quantity function is linear:
   \[
   q(p) = 4000 - 400(p - 5)
   \]
   Simplifying:
   \[
   q(p) = 6000 - 400p
   \]

4. **Cost Function:**
   The cost to the manufacturer is $2 per shirt. So the total cost $ C(p) $ for selling $ q(p) $ shirts is:
   \[
   C(p) = 2 \cdot q(p) = 2(6000 - 400p) = 12000 - 800p
   \]

5. **Profit Function:**
   The profit is the revenue minus the cost. So, the profit function $ \Pi(p) $ is:
   \[
   \Pi(p) = R(p) - C(p)
   \]
   Substituting $ R(p) = p \cdot (6000 - 400p) $ and $ C(p) = 12000 - 800p $:
   \[
   \Pi(p) = p(6000 - 400p) - (12000 - 800p)
   \]
   Simplifying:
   \[
   \Pi(p) = 6000p - 400p^2 - 12000 + 800p
   \]
   \[
   \Pi(p) = -400p^2 + 6800p - 12000
   \]

6. **Maximizing Profit:**
   To maximize the profit, we take the derivative of $ \Pi(p) $ with respect to $ p $ and set it equal to zero:
   \[
   \frac{d\Pi(p)}{dp} = -800p + 6800 = 0
   \]
   Solving for $ p $:
   \[
   800p = 6800
   \]
   \[
   p = \frac{6800}{800} = 8.5
   \]

So, the manufacturer should sell the T-shirts at **$8.50** to maximize profit.

Would you like a deeper explanation of any step, or do you have further questions?

### Related Questions:
1. What happens to the number of shirts sold at the optimal price of $8.50?
2. How would the profit function change if the cost per shirt increased to $3?
3. How sensitive is the profit to changes in the price around $8.50?
4. What is the maximum possible profit the manufacturer can achieve?
5. How would the result change if for each $1 increase in price, 500 fewer shirts were sold?

**Tip:** When maximizing profit, it's crucial to consider both the revenue and the cost functions carefully to ensure you're accounting for all factors affecting profitability.

A manufacturer can produce souvenir T-shirts at a cost of $2 apiece. The shirts have been selling for $5 apiece, and at this price, tourists have been buying 4000 shirts a month. The manufacturer is planning to raise the price of the shirts and estimates that for each $1 increase in the price, 400 fewer shirts will be sold each month. At what price should the manufacturer sell the shirts to maximize profit?

Learn how to solve a profit maximization problem involving souvenir T-shirts using quadratic equations. This problem focuses on algebraic concepts such as revenue, cost, and profit functions, and is suitable for high school students in Grades 9-12.

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