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Ch. 3 - Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 3, Problem 22b

Economics


Marginal revenue
Suppose that the revenue from selling x washing machines is


r(x) = 20000(1 − 1/x) dollars.


b. Use the function r'(x) to estimate the increase in revenue that will result from increasing production from 100 machines a week to 101 machines a week.

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1
First, understand that the function r(x) = 20000(1 - 1/x) represents the revenue from selling x washing machines. We need to find the derivative r'(x) to determine the marginal revenue, which is the rate of change of revenue with respect to the number of machines sold.
To find r'(x), apply the derivative rules. The function r(x) = 20000(1 - 1/x) can be rewritten as r(x) = 20000 - 20000/x. Differentiate this function with respect to x.
The derivative of a constant is zero, so the derivative of 20000 is 0. For the term -20000/x, use the power rule. Rewrite -20000/x as -20000x^(-1) and differentiate to get 20000x^(-2) or 20000/x^2.
Thus, the derivative r'(x) = 20000/x^2. This represents the marginal revenue function, which gives the approximate change in revenue for a small change in the number of machines sold.
To estimate the increase in revenue from increasing production from 100 machines to 101 machines, evaluate r'(100). Substitute x = 100 into the marginal revenue function: r'(100) = 20000/(100^2). This value will give the estimated increase in revenue for producing one additional machine.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Derivative

The derivative of a function measures how the function's output changes as its input changes. In this context, r'(x) represents the rate of change of revenue with respect to the number of washing machines sold. Calculating r'(x) allows us to estimate the marginal revenue, which is the additional revenue generated by selling one more unit.
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Marginal Revenue

Marginal revenue is the additional income from selling one more unit of a good or service. It is calculated as the derivative of the revenue function with respect to quantity. In this problem, r'(x) gives the marginal revenue, helping estimate the revenue increase when production rises from 100 to 101 machines.
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Function Evaluation

Function evaluation involves substituting specific values into a function to determine its output. Here, evaluating r'(x) at x = 100 provides the marginal revenue for the 100th machine, which helps estimate the revenue change when production increases to 101 machines. This step is crucial for applying the derivative to real-world scenarios.
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