The latest demand equation for your Yoda vs. Alien T-shirts is given by
q = −50x + 700
where q is the number of shirts you can sell in one week if you charge $x per shirt. The Student Council charges you $100 per week for use of their facilities, and the T-shirts cost you $8 each. Find the weekly cost as a function of the unit price x.
C(x) =

Hence, find the weekly profit as a function of x.
P(x) =

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

To solve this problem, we need to find the weekly cost, profit, and the unit price $ x $ that maximizes profit for the Yoda vs. Alien T-shirts.

### 1. Weekly Cost as a Function of $ x $

The cost consists of two parts:
- A fixed cost of $100 for the use of facilities.
- A variable cost of $8 per T-shirt produced.

The number of T-shirts sold, $ q $, is given by the demand equation:
\[
q = -50x + 700
\]
So, the weekly cost function $ C(x) $ includes the cost of producing $ q $ T-shirts at $8 each, plus the fixed cost of $100:
\[
C(x) = 8q + 100 = 8(-50x + 700) + 100
\]
Simplifying:
\[
C(x) = 8(-50x + 700) + 100
C(x) = -400x + 5600 + 100
C(x) = -400x + 5700
\]
Thus, the weekly cost function is:
\[
C(x) = -400x + 5700
\]

### 2. Weekly Profit as a Function of $ x $

The profit function $ P(x) $ is given by the revenue minus the cost. 

The revenue is the product of the price $ x $ and the quantity $ q $ sold:
\[
\text{Revenue} = xq = x(-50x + 700) = -50x^2 + 700x
\]
Now, the profit function $ P(x) $ is:
\[
P(x) = \text{Revenue} - \text{Cost}
P(x) = (-50x^2 + 700x) - (-400x + 5700)
\]
Simplifying:
\[
P(x) = -50x^2 + 700x + 400x - 5700
P(x) = -50x^2 + 1100x - 5700
\]
Thus, the weekly profit function is:
\[
P(x) = -50x^2 + 1100x - 5700
\]

### 3. Maximizing the Weekly Profit

To maximize the profit, we take the derivative of $ P(x) $ with respect to $ x $, set it equal to zero, and solve for $ x $.

The derivative of $ P(x) $ is:
\[
P'(x) = -100x + 1100
\]
Setting $ P'(x) = 0 $:
\[
-100x + 1100 = 0
\]
Solving for $ x $:
\[
x = \frac{1100}{100} = 11
\]
So, the unit price that maximizes profit is $ x = 11 $ dollars per T-shirt.

### 4. Largest Possible Weekly Profit

To find the largest possible weekly profit, we substitute $ x = 11 $ into the profit function $ P(x) $:
\[
P(11) = -50(11)^2 + 1100(11) - 5700
\]
Simplifying:
\[
P(11) = -50(121) + 12100 - 5700
P(11) = -6050 + 12100 - 5700
P(11) = 350
\]
Thus, the largest possible weekly profit is **$350**.

### Summary:
- **Cost function**: $ C(x) = -400x + 5700 $
- **Profit function**: $ P(x) = -50x^2 + 1100x - 5700 $
- **Optimal unit price**: $ x = 11 $ dollars per T-shirt
- **Maximum profit**: $ 350 $ dollars per week

Would you like more details on any part of this solution? Here are 5 related questions to explore further:

1. How would a change in fixed costs affect the profit-maximizing price?
2. What happens to profit if the T-shirt cost increases from $8 to $10?
3. Can you explain how the derivative helps in maximizing profit?
4. What is the impact on the profit function if the demand changes to $ q = -60x + 800 $?
5. How sensitive is the maximum profit to changes in the unit price $ x $?

**Tip**: When maximizing profit, always check that the second derivative is negative to ensure you have a maximum, not a minimum.

The latest demand equation for your Yoda vs. Alien T-shirts is given by q = −50x + 700 where q is the number of shirts you can sell in one week if you charge $x per shirt. The Student Council charges you $100 per week for use of their facilities, and the T-shirts cost you $8 each. Find the weekly cost as a function of the unit price x. C(x) = Hence, find the weekly profit as a function of x. P(x) = Determine the unit price (in dollars per T-shirt) you should charge to obtain the largest possible weekly profit. x = $ per T-shirt What is the largest possible weekly profit (in dollars)? $

Learn how to calculate the optimal price and maximum profit for Yoda vs. Alien T-shirts using demand and cost functions. This detailed step-by-step solution covers key algebraic concepts, including the demand equation, profit function, and optimization using derivatives. Suitable for students in Grades 10-12.

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