Bakers' Club is trying to raise funds by selling premium chocolate chip cookies in a school fair. The variable cost to make each cookie is $15.00 and it is being sold for P25.00. So far, the organization has already shelled out #790.00 for the cookie sale. Find the profit function P(x) where x represents the number of cookies sold. Hint: Profit Total Revenue - Total Cost Total Revenue = Price per unit x quantity sold Total Cost = Total variable cost + fixed cost b. If 146 cookies were sold, how much is the total profit? c. How many cookies must be made and sold to break even? Hint: Break even point is the zero of P(x) d. How many cookie should be sold to gain a profit of P250.00?

Let's break down the problem step by step.

### Part a: Finding the Profit Function $ P(x) $

To find the profit function $ P(x) $, we use the following formulas:

- **Total Revenue (TR):** This is the total amount earned from selling the cookies.
  \[
  \text{Total Revenue} = \text{Price per unit} \times \text{quantity sold}
  \]
  Given that each cookie is sold for $ P25.00 $, the total revenue is:
  \[
  \text{TR}(x) = 25x
  \]

- **Total Cost (TC):** This is the sum of the fixed costs and the total variable cost.
  \[
  \text{Total Cost} = \text{Fixed Cost} + \text{Total Variable Cost}
  \]
  The fixed cost is given as $ P790.00 $. The total variable cost is the cost to make each cookie multiplied by the number of cookies sold, so:
  \[
  \text{Total Variable Cost} = 15x
  \]
  Therefore, the total cost is:
  \[
  \text{TC}(x) = 790 + 15x
  \]

- **Profit Function (P(x)):** Profit is the difference between total revenue and total cost.
  \[
  P(x) = \text{TR}(x) - \text{TC}(x)
  \]
  Substituting the expressions we have:
  \[
  P(x) = 25x - (790 + 15x)
  \]
  Simplifying this:
  \[
  P(x) = 25x - 790 - 15x = 10x - 790
  \]

### Part b: Total Profit for 146 Cookies Sold

To find the profit when 146 cookies are sold, substitute $ x = 146 $ into the profit function $ P(x) $:

\[
P(146) = 10(146) - 790
\]
\[
P(146) = 1460 - 790 = 670
\]

So, the total profit when 146 cookies are sold is $ P670.00 $.

### Part c: Break-even Point

The break-even point is when the profit is zero, $ P(x) = 0 $. Set the profit function equal to zero and solve for $ x $:

\[
0 = 10x - 790
\]
\[
10x = 790
\]
\[
x = \frac{790}{10} = 79
\]

So, 79 cookies must be sold to break even.

### Part d: Number of Cookies to Gain a Profit of P250.00

To find the number of cookies needed to achieve a profit of $ P250.00 $, set $ P(x) = 250 $ and solve for $ x $:

\[
250 = 10x - 790
\]
\[
10x = 250 + 790 = 1040
\]
\[
x = \frac{1040}{10} = 104
\]

So, 104 cookies need to be sold to gain a profit of $ P250.00 $.

---

Would you like further details or have any questions?

Here are some related questions:
1. What is the meaning of the break-even point in business?
2. How would the profit function change if the selling price increased?
3. What is the impact of a higher fixed cost on the break-even point?
4. How could the Bakers' Club maximize its profit?
5. What if the variable cost increases to P20.00 per cookie, how would that affect the profit function?

**Tip:** Always double-check your fixed and variable costs when calculating the break-even point.

Bakers' Club is trying to raise funds by selling premium chocolate chip cookies in a school fair. The variable cost to make each cookie is $15.00 and it is being sold for P25.00. So far, the organization has already shelled out $790.00 for the cookie sale. Find the profit function P(x) where x represents the number of cookies sold. Hint: Profit = Total Revenue - Total Cost. Total Revenue = Price per unit x quantity sold. Total Cost = Total variable cost + fixed cost. b. If 146 cookies were sold, how much is the total profit? c. How many cookies must be made and sold to break even? Hint: Break-even point is the zero of P(x). d. How many cookies should be sold to gain a profit of P250.00?

Learn how to calculate the profit function for cookie sales, determine the break-even point, and analyze total profit for specific quantities. This problem involves key concepts in linear functions and profit analysis, suitable for students in Grades 10-12.

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