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?

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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.