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Let’s analyze each part of this problem.

### Given:
\( r(x) = 20,000 \left(1 - \frac{1}{x}\right) \)

where \( r(x) \) represents the revenue from selling \( x \) washing machines.

#### **Part a**: Find the marginal revenue when 100 machines are produced.

The marginal revenue is given by the derivative \( r'(x) \), which represents the rate of change of revenue with respect to the number of washing machines. We need to differentiate \( r(x) \) with respect to \( x \).

1. **Differentiate** \( r(x) \):
   \[
   r'(x) = \frac{d}{dx} \left[ 20,000 \left(1 - \frac{1}{x}\right) \right] 
   \]
   \[
   r'(x) = 20,000 \cdot \frac{d}{dx} \left(1 - \frac{1}{x}\right) 
   \]
   \[
   r'(x) = 20,000 \left(0 + \frac{1}{x^2}\right) 
   \]
   \[
   r'(x) = \frac{20,000}{x^2}
   \]

2. **Evaluate** \( r'(x) \) at \( x = 100 \):
   \[
   r'(100) = \frac{20,000}{100^2} = \frac{20,000}{10,000} = 2
   \]

Thus, the marginal revenue when 100 machines are produced is **2 dollars** per additional washing machine.

#### **Part b**: Estimate the increase in revenue from increasing production from 100 machines to 101 machines.

Since the marginal revenue \( r'(x) \) represents the approximate increase in revenue for one additional machine, the increase in revenue from 100 to 101 machines can be approximated by \( r'(100) \). Therefore, the increase is approximately **2 dollars**.

#### **Part c**: Find the limit of \( r'(x) \) as \( x \) approaches infinity. How would you interpret this number?

We need to find:
\[
\lim_{x \to \infty} r'(x) = \lim_{x \to \infty} \frac{20,000}{x^2}
\]

As \( x \) approaches infinity, \( x^2 \) grows much faster than the constant 20,000, so the fraction \( \frac{20,000}{x^2} \) approaches zero. Therefore:
\[
\lim_{x \to \infty} r'(x) = 0
\]

### Interpretation:
The limit approaching zero indicates that as the number of washing machines produced becomes very large, the additional revenue gained from producing one more washing machine approaches zero. This suggests diminishing marginal returns as production increases.

### Summary:
- **a.** \( r'(100) = 2 \) dollars
- **b.** The estimated increase in revenue from 100 to 101 machines is **2 dollars**.
- **c.** \( \lim_{x \to \infty} r'(x) = 0 \), which indicates diminishing marginal returns with increasing production.

Would you like further clarification on any of these parts or additional details?

Here are 5 related questions:
1. How is marginal revenue conceptually different from total revenue?
2. What would happen to the marginal revenue if the production costs increased?
3. How does the behavior of the derivative reflect diminishing returns in this context?
4. What is the economic significance of a zero marginal revenue as production increases?
5. How could marginal revenue analysis inform decisions about optimal production levels?

**Tip:** Remember that derivatives provide a way to understand the rate of change, helping economists and business owners make informed decisions based on changing conditions.

Solve problem 22 involving marginal revenue where the revenue function is r(x) = 20,000(1 - 1/x) dollars.

Learn how to calculate the marginal revenue when producing washing machines with a revenue function r(x) = 20,000(1 - 1/x). Includes detailed solutions for finding r'(100), estimating revenue changes, and analyzing the limit of r'(x) as x approaches infinity.

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