T-Shirt Profit   Two fraternities, Sig Ep and Ep Sig, plan to raise money jointly to benefit homeless people on Long Island. They will sell Yoda vs. Alien T-shirts in the student center, but are not sure how much to charge. Sig Ep treasurer Augustus recalls that they once sold 120 shirts in a week at $5 per shirt, but Ep Sig treasurer Julius has solid research indicating that it is possible to sell 200 per week at $3 per shirt.
(a)
On the basis of this information, construct a linear demand equation for Yoda vs. Alien T-shirts. Hence obtain the weekly revenue R as a function of the unit price x.
R(x) = 
 
(b)
The university administration charges the fraternities a weekly fee of $200 for use of the student center. Write down the weekly profit P as a function of the unit price x.
P(x) = 
 
Determine how much the fraternities should charge (in dollars per T-shirt) to obtain the largest possible weekly profit.
x = $ 
 per T-shirt
What is the largest possible weekly profit (in dollars)?
$ 


Let's break down the problem step-by-step to construct the demand equation, revenue function, profit function, and find the optimal price for maximum profit.

### (a) Demand Equation and Revenue Function

We have two price-demand points given:

1. **Point 1**: At a price of $5 per shirt, they sold 120 shirts. This gives the point $(x_1, q_1) = (5, 120)$.
2. **Point 2**: At a price of $3 per shirt, they sold 200 shirts. This gives the point $(x_2, q_2) = (3, 200)$.

We need to find the linear demand equation of the form:

\[
q = mx + b
\]

where $ q $ is the quantity sold, $ x $ is the price, and $ m $ and $ b $ are constants.

#### Step 1: Find the slope $ m $

The slope $ m $ is given by:

\[
m = \frac{q_2 - q_1}{x_2 - x_1} = \frac{200 - 120}{3 - 5} = \frac{80}{-2} = -40
\]

#### Step 2: Find the intercept $ b $

Using the point-slope form, we use one of the points, say $(5, 120)$:

\[
q - 120 = -40(x - 5)
\]

Simplifying:

\[
q - 120 = -40x + 200
\]

\[
q = -40x + 320
\]

So, the demand equation is:

\[
q = -40x + 320
\]

#### Step 3: Revenue Function $ R(x) $

The revenue $ R(x) $ is given by the product of the price per shirt and the quantity sold:

\[
R(x) = x \cdot q = x \cdot (-40x + 320)
\]

\[
R(x) = -40x^2 + 320x
\]

### (b) Profit Function

The weekly profit $ P(x) $ is the revenue minus the costs. The costs include the fixed weekly fee of $200 for using the student center.

\[
P(x) = R(x) - 200
\]

\[
P(x) = -40x^2 + 320x - 200
\]

### (c) Optimal Price for Maximum Profit

To find the price $ x $ that maximizes the profit, we need to find the critical points of $ P(x) $ by taking its derivative and setting it to zero.

#### Step 1: Derivative of $ P(x) $

\[
P'(x) = \frac{d}{dx} \left( -40x^2 + 320x - 200 \right)
\]

\[
P'(x) = -80x + 320
\]

#### Step 2: Set $ P'(x) = 0 $ to find the critical point

\[
-80x + 320 = 0
\]

\[
-80x = -320
\]

\[
x = 4
\]

So, the fraternities should charge **$4 per T-shirt** to maximize their profit.

#### Step 3: Maximum Profit

Substitute $ x = 4 $ back into the profit function:

\[
P(4) = -40(4)^2 + 320(4) - 200
\]

\[
P(4) = -40(16) + 1280 - 200
\]

\[
P(4) = -640 + 1280 - 200
\]

\[
P(4) = 440
\]

Thus, the largest possible weekly profit is **$440**.

### Summary

1. **Demand equation**: $ q = -40x + 320 $
2. **Revenue function**: $ R(x) = -40x^2 + 320x $
3. **Profit function**: $ P(x) = -40x^2 + 320x - 200 $
4. **Optimal price**: $ x = 4 $ dollars per T-shirt
5. **Maximum profit**: $440

Would you like more details on any part of the solution?

---

**5 Related Questions:**

1. How does the demand equation change if they decide to change the price range?
2. What would be the effect on the profit function if the university fee increased to $300?
3. How would a decrease in demand at lower prices affect the revenue function?
4. How can we modify the profit function to include a variable cost per shirt?
5. What would be the maximum revenue, and at what price would it occur?

**Tip**: The maximum profit always occurs at a critical point where the derivative of the profit function equals zero.

T-Shirt Profit   Two fraternities, Sig Ep and Ep Sig, plan to raise money jointly to benefit homeless people on Long Island. They will sell Yoda vs. Alien T-shirts in the student center, but are not sure how much to charge. Sig Ep treasurer Augustus recalls that they once sold 120 shirts in a week at $5 per shirt, but Ep Sig treasurer Julius has solid research indicating that it is possible to sell 200 per week at $3 per shirt.
(a)
On the basis of this information, construct a linear demand equation for Yoda vs. Alien T-shirts. Hence obtain the weekly revenue R as a function of the unit price x.
R(x) =

(b)
The university administration charges the fraternities a weekly fee of $200 for use of the student center. Write down the weekly profit P as a function of the unit price x.
P(x) =

Determine how much the fraternities should charge (in dollars per T-shirt) to obtain the largest possible weekly profit.
x = $ per T-shirt
What is the largest possible weekly profit (in dollars)?

Learn how to construct a linear demand equation and maximize weekly profit for a T-shirt sale. This problem involves creating a revenue function, applying profit equations, and solving for optimal prices using calculus techniques. Suitable for Grades 10-12 students.

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