2 worded problem involving rational function, equation and inequalities. Show your complete solution to your problem

### Word Problem Involving Rational Functions, Equations, and Inequalities

**Problem:**

A certain company's cost of production $ C(x) $, in dollars, to produce $ x $ units of a product is given by the function:

\[
C(x) = \frac{5000 + 3x}{x - 2}
\]

where $ x $ represents the number of units produced, and $ x > 2 $. 

1. **Question 1:** Find the number of units $ x $ that must be produced so that the cost per unit is less than $20. 
2. **Question 2:** Determine the value of $ x $ for which the cost of production $ C(x) $ equals $50.

---

### Solution:

#### 1. **Finding the number of units $ x $ so that $ C(x) < 20 $:**

We need to solve the inequality:

\[
\frac{5000 + 3x}{x - 2} < 20
\]

Let's start by solving this step by step.

1. **Isolate the rational function:**

\[
\frac{5000 + 3x}{x - 2} < 20
\]

2. **Subtract 20 from both sides:**

\[
\frac{5000 + 3x}{x - 2} - 20 < 0
\]

3. **Combine the terms under a common denominator:**

\[
\frac{5000 + 3x - 20(x - 2)}{x - 2} < 0
\]

4. **Simplify the numerator:**

\[
\frac{5000 + 3x - 20x + 40}{x - 2} < 0
\]

\[
\frac{5040 - 17x}{x - 2} < 0
\]

5. **Analyze the inequality:**

   To find where this rational expression is negative, determine the critical points by setting the numerator and denominator equal to zero:

   - Numerator: $ 5040 - 17x = 0 $ 
   - Denominator: $ x - 2 = 0 $

   Solving these:

   \[
   5040 - 17x = 0 \quad \Rightarrow \quad x = \frac{5040}{17} \approx 296.47
   \]

   \[
   x - 2 = 0 \quad \Rightarrow \quad x = 2
   \]

6. **Sign Analysis:**

   Perform a sign analysis by considering intervals around these critical points, $ x = 2 $ and $ x = 296.47 $.

   - For $ x < 2 $: The expression is undefined (since the denominator becomes zero).
   - For $ 2 < x < 296.47 $: The expression $ \frac{5040 - 17x}{x - 2} $ is positive.
   - For $ x > 296.47 $: The expression $ \frac{5040 - 17x}{x - 2} $ is negative.

   Thus, $ C(x) < 20 $ when $ x > 296.47 $.

#### **Answer:** To ensure that the cost per unit is less than $20, more than 297 units must be produced.

---

#### 2. **Determine the value of $ x $ for which $ C(x) = 50 $:**

We now solve the equation:

\[
\frac{5000 + 3x}{x - 2} = 50
\]

1. **Isolate the rational function:**

\[
\frac{5000 + 3x}{x - 2} = 50
\]

2. **Cross multiply to eliminate the fraction:**

\[
5000 + 3x = 50(x - 2)
\]

3. **Distribute and simplify:**

\[
5000 + 3x = 50x - 100
\]

\[
5000 + 100 = 50x - 3x
\]

\[
5100 = 47x
\]

4. **Solve for $ x $:**

\[
x = \frac{5100}{47} \approx 108.51
\]

#### **Answer:** The cost of production is $50 when approximately 109 units are produced.

---

### Summary:
1. **To have a cost per unit less than $20, more than 297 units must be produced.**
2. **The cost of production equals $50 when approximately 109 units are produced.**

---

Would you like further details or have any questions?

### Related Questions:
1. How do you find the domain of the cost function $ C(x) = \frac{5000 + 3x}{x - 2} $?
2. What happens to the cost $ C(x) $ as $ x $ approaches infinity?
3. How would the problem change if the function had an additional fixed cost?
4. Can you graph $ C(x) $ and identify any asymptotes?
5. How does the cost function behave as $ x $ approaches 2 from the right?

---

### Tip:
When solving rational inequalities, always perform a sign analysis by testing values in the critical intervals determined by the roots of the numerator and denominator.

Explore the solution to a word problem involving rational functions, equations, and inequalities. Learn how to calculate production costs using the formula C(x) = (5000 + 3x) / (x - 2) for a certain company's production units. Suitable for advanced high school or college-level mathematics.

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