Suppose a company has fixed costs of $45,000 and variable cost per unit of
4
9
x + 222 dollars,
where x is the total number of units produced. Suppose further that the selling price of its product is
1752 −
5
9
x dollars per unit.
(a) Find the simplified cost function C(x.

C(x) =

(b) Find the simplified revenue function R(x).

R(x) =

(c) Form the profit function P(x).

P(x) =

Let's break down the problem and solve each part step by step.

### Part (a): Find the simplified cost function $ C(x) $.

The **cost function** $ C(x) $ is the sum of the **fixed costs** and the **variable costs**. The fixed cost is given as $45,000. The variable cost per unit is $ \frac{4}{9}x + 222 $ dollars.

Thus, the total cost function $ C(x) $ is:
\[
C(x) = \text{Fixed Costs} + \text{Variable Costs}
\]
\[
C(x) = 45000 + x \times \left( \frac{4}{9}x + 222 \right)
\]

Simplifying the expression:
\[
C(x) = 45000 + \left( \frac{4}{9}x^2 + 222x \right)
\]
\[
C(x) = \frac{4}{9}x^2 + 222x + 45000
\]

So, the simplified cost function is:
\[
C(x) = \frac{4}{9}x^2 + 222x + 45000
\]

---

### Part (b): Find the simplified revenue function $ R(x) $.

The **revenue function** $ R(x) $ is the selling price per unit times the number of units produced, $ x $. The selling price per unit is $ 1752 - \frac{5}{9}x $.

Thus, the revenue function $ R(x) $ is:
\[
R(x) = x \times \left( 1752 - \frac{5}{9}x \right)
\]
Simplifying the expression:
\[
R(x) = 1752x - \frac{5}{9}x^2
\]

So, the simplified revenue function is:
\[
R(x) = 1752x - \frac{5}{9}x^2
\]

---

### Part (c): Form the profit function $ P(x) $.

The **profit function** $ P(x) $ is the difference between the revenue and the cost functions:
\[
P(x) = R(x) - C(x)
\]
Substitute the expressions for $ R(x) $ and $ C(x) $:
\[
P(x) = \left( 1752x - \frac{5}{9}x^2 \right) - \left( \frac{4}{9}x^2 + 222x + 45000 \right)
\]
Simplifying:
\[
P(x) = 1752x - \frac{5}{9}x^2 - \frac{4}{9}x^2 - 222x - 45000
\]
\[
P(x) = 1752x - 222x - \left( \frac{5}{9}x^2 + \frac{4}{9}x^2 \right) - 45000
\]
\[
P(x) = 1530x - \frac{9}{9}x^2 - 45000
\]
\[
P(x) = 1530x - x^2 - 45000
\]

Thus, the simplified profit function is:
\[
P(x) = 1530x - x^2 - 45000
\]

---

Would you like more details on any of these steps or have any specific questions? Here are some related questions:

1. How do fixed and variable costs impact the cost function?
2. What happens to the revenue if the number of units produced increases significantly?
3. How can the company maximize its profit given the profit function $ P(x) $?
4. What does the quadratic term in the profit function suggest about the relationship between production and profit?
5. How would changing the fixed costs affect the cost function?

**Tip:** Maximizing the profit function can be done by finding the derivative of $ P(x) $ and setting it equal to zero to solve for $ x $. This will give the number of units that maximize profit.

Suppose a company has fixed costs of $45,000 and variable cost per unit of 49x + 222 dollars, where x is the total number of units produced. Suppose further that the selling price of its product is 1752 − 59x dollars per unit.
(a) Find the simplified cost function C(x).
(b) Find the simplified revenue function R(x).
(c) Form the profit function P(x).

Learn how to simplify cost, revenue, and profit functions for a company with fixed costs of $45,000 and variable costs per unit given by 49x + 222. Explore key concepts in algebra and quadratic functions, including cost, revenue, and profit calculation. Suitable for high school students in Grades 10-12.

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