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 70 shirts in a week at $5 per shirt, but Ep Sig treasurer Julius has solid research indicating that it is possible to sell 280 per week at $2 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 $100 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 solve this step by step.

### (a) Linear Demand Equation

We are given two data points regarding the number of T-shirts sold at different prices:
- 70 shirts sold at $ x = 5 $ dollars per shirt.
- 280 shirts sold at $ x = 2 $ dollars per shirt.

Let $ q $ represent the quantity of T-shirts sold per week, and $ x $ represent the price per T-shirt. The demand equation is typically a linear relationship between $ q $ and $ x $, of the form:
\[
q(x) = mx + b
\]
We need to find the slope $ m $ and intercept $ b $ using the two points $(5, 70)$ and $(2, 280)$.

The slope $ m $ of the line is given by the formula:
\[
m = \frac{q_2 - q_1}{x_2 - x_1} = \frac{280 - 70}{2 - 5} = \frac{210}{-3} = -70
\]
Thus, the slope $ m = -70 $.

Now we use the point-slope form of the equation to find the intercept $ b $. Using the point $ (x_1 = 5, q_1 = 70) $, we plug into the equation $ q = mx + b $:
\[
70 = -70(5) + b
\]
\[
70 = -350 + b
\]
\[
b = 420
\]
Thus, the demand equation is:
\[
q(x) = -70x + 420
\]

### Revenue Equation

Revenue $ R(x) $ is the product of the price per T-shirt $ x $ and the number of T-shirts sold $ q(x) $. Therefore:
\[
R(x) = x \cdot q(x) = x(-70x + 420)
\]
\[
R(x) = -70x^2 + 420x
\]

### (b) Profit Equation

We are told that the university administration charges a weekly fee of $100, so the weekly profit $ P(x) $ is the revenue $ R(x) $ minus the fee. Therefore:
\[
P(x) = R(x) - 100 = (-70x^2 + 420x) - 100
\]
\[
P(x) = -70x^2 + 420x - 100
\]

### Maximizing the Profit

To find the price $ x $ that maximizes profit, we take the derivative of $ P(x) $ with respect to $ x $ and set it equal to zero.

\[
P'(x) = \frac{d}{dx} \left( -70x^2 + 420x - 100 \right)
\]
\[
P'(x) = -140x + 420
\]
Set $ P'(x) = 0 $ to find the critical points:
\[
-140x + 420 = 0
\]
\[
140x = 420
\]
\[
x = \frac{420}{140} = 3
\]

Thus, the price per T-shirt that maximizes profit is $ x = 3 $ dollars.

### Largest Possible Weekly Profit

To find the maximum profit, substitute $ x = 3 $ into the profit equation $ P(x) $:
\[
P(3) = -70(3)^2 + 420(3) - 100
\]
\[
P(3) = -70(9) + 1260 - 100
\]
\[
P(3) = -630 + 1260 - 100
\]
\[
P(3) = 530
\]

Thus, the largest possible weekly profit is **$530**.

### Final Answer
- The fraternities should charge **$3** per T-shirt.
- The largest possible weekly profit is **$530**.

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

Here are 5 related questions:
1. How does changing the price of T-shirts affect the demand?
2. What would happen to the maximum profit if the fee increased?
3. How can you graph the profit function to visualize the maximum?
4. What other pricing strategies could be considered for maximizing revenue?
5. What is the significance of the slope in the demand equation?

**Tip:** To find maxima or minima in profit functions, always take the derivative and look for critical points where the derivative equals zero.

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 70 shirts in a week at $5 per shirt, but Ep Sig treasurer Julius has solid research indicating that it is possible to sell 280 per week at $2 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 $100 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 maximize the profit for T-shirt sales by constructing a linear demand equation, calculating the weekly revenue function, and determining the optimal price per T-shirt. This problem covers linear equations, profit functions, and the use of derivatives for finding maximum profit. Suitable for high school students in Grades 10-12.

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