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Marginal revenue Suppose that the revenue from selling \(x\) washing machines is $$ r(x)=20,000\left(1-\frac{1}{x}\right) $$ dollars. a. Find the marginal revenue when 100 machines are produced. b. Use the function \(r^{\prime}(x)\) to estimate the increase in revenue that will result from increasing production from 100 machines a week to 101 machines a week. c. Find the limit of \(r^{\prime}(x)\) as \(x \rightarrow \infty .\) How would you interpret this number?

Short Answer

Expert verified
a) Marginal revenue at 100 machines is 2 dollars. b) Revenue increase from 100 to 101 machines is approximately 2 dollars. c) Limit is 0; marginal revenue diminishes as production increases.

Step by step solution

01

Find the Derivative

First, let's find the derivative of the revenue function, which will give us the marginal revenue function. The given function is \( r(x) = 20,000 \left(1 - \frac{1}{x} \right) \). We use the power rule and the chain rule to differentiate this function.The derivative \( r'(x) \) is:\[ r'(x) = -20,000 \times \left(-\frac{1}{x^2}\right) = \frac{20,000}{x^2}\].
02

Calculate Marginal Revenue at x=100

Now, substitute \( x = 100 \) into the derivative \( r'(x) \) to find the marginal revenue when 100 machines are produced.\[ r'(100) = \frac{20,000}{100^2} = \frac{20,000}{10,000} = 2\]The marginal revenue when 100 machines are produced is 2 dollars.
03

Estimate Revenue Increase from 100 to 101 Machines

To estimate the increase in revenue when production increases from 100 to 101 machines, we use the marginal revenue \( r'(100) \). Since \( r'(100) = 2 \), which means the revenue increases by approximately 2 dollars when production increases by 1 washing machine.Therefore, the revenue increase is approximately 2 dollars.
04

Find Limit of r'(x) as x Approaches Infinity

Finally, calculate the limit of \( r'(x) \) as \( x \to \infty \). We analyze the function \( r'(x) = \frac{20,000}{x^2} \).As \( x \to \infty \), \( x^2 \to \infty \), and thus \( \frac{20,000}{x^2} \to 0 \).So, \( \lim_{x \to \infty} r'(x) = 0 \).
05

Interpret the Limit

The interpretation of \( \lim_{x \to \infty} r'(x) = 0 \) is that as the number of washing machines sold increases indefinitely, the additional revenue gained by selling an additional machine approaches zero. This implies that after a certain point, increasing production offers diminishing returns in terms of revenue.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Derivative in Marginal Revenue
To grasp the concept of marginal revenue, it's essential to understand derivatives first. The derivative of a function gives you the rate at which the function's value is changing at any given point. For revenue functions, this means it tells you how the revenue changes as you sell one more unit.

In our exercise, the revenue function is expressed as \[ r(x) = 20,000 imes igg(1 - \frac{1}{x}\bigg) \]. By differentiating this function, we find the derivative, which is called the marginal revenue function. In this case, \[ r'(x) = \frac{20,000}{x^2} \].

This derivative informs us how the revenue changes with respect to an increase in the number of washing machines sold. Essentially, it allows businesses to make informed decisions about how many additional units to produce.
Calculating Revenue Increase
Calculating revenue increase involves understanding how much extra revenue one more unit sold will bring. The marginal revenue calculated at any specific quantity provides this estimate.

For instance, from the original exercise, when 100 machines are sold, the marginal revenue is \[ r'(100) = 2 \], indicating that producing and selling one additional washing machine will increase revenue by approximately 2 dollars. This calculation helps businesses determine the benefit of scaling up production slightly, especially in the short term where only one or a few units are added.

  • Helps in decision-making for production increases.
  • Assists in financial forecasting and budgeting.
Interpreting Limits in Revenue Functions
The limit of a function helps us understand its behavior as the input approaches a certain value. In the context of marginal revenue, the limit as the number of units approaches infinity (a very large number) tells us about long-term trends.

For the revenue function's derivative \[ r'(x) = \frac{20,000}{x^2} \], the limit as \( x \to \infty \) is \( 0 \). This result indicates that as more and more machines are sold, the additional revenue from each machine decreases and eventually approaches zero.

  • Expected in saturated markets.
  • Important for long-term strategic planning.
Exploring Diminishing Returns
Diminishing returns is a principal concept in economics referring to a point where adding an additional factor of production results in smaller increases in output. In the context of marginal revenue, it means that as you continue producing more units, the added revenue from each additional unit declines.

From our exercise, when \( \lim_{x \to \infty} r'(x) = 0 \), we interpret this to mean that selling more washing machines results in lesser additional revenue per unit. This phenomenon occurs due to market saturation where all potential customers already have the product, or because of the cost structures that increase with production.

  • Critical for maximizing profit efficiency.
  • Guides on optimal production quantity.

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