Question

Suppose that the revenue \(R(n)\) in dollars from producing \(n\) computers is given by \(R(n)=0.4 n-0.001 n^{2} .\) Find the instantaneous rates of change of revenue when \(n=10\) and \(n=100\). (The instantaneous rate of change of revenue with respect to the amount of product produced is called the marginal revenue.)

Short answer

Marginal revenue is 0.38 dollars when \( n=10 \) and 0.2 dollars when \( n=100 \).

Step-by-step solution

Understand the Problem

We are given the revenue function \( R(n) = 0.4n - 0.001n^2 \). We need to find the instantaneous rate of change of revenue (marginal revenue) at specific production levels \( n = 10 \) and \( n = 100 \). This involves finding the derivative of the revenue function and then evaluating it at these specific values of \( n \).

Differentiate the Revenue Function

First, we'll compute the derivative of the revenue function \( R(n) = 0.4n - 0.001n^2 \). The derivative \( R'(n) \) is found using basic differentiation rules:- The derivative of \( 0.4n \) is \( 0.4 \).- The derivative of \( -0.001n^2 \) is \( -0.002n \).Thus, \( R'(n) = 0.4 - 0.002n \).

Evaluate the Derivative at \( n = 10 \)

Now, substitute \( n = 10 \) into the derivative to find the marginal revenue when \( n = 10 \):\[R'(10) = 0.4 - 0.002 \times 10 = 0.4 - 0.02 = 0.38\]Therefore, the marginal revenue when \( n = 10 \) is 0.38 dollars per computer.

Evaluate the Derivative at \( n = 100 \)

Next, substitute \( n = 100 \) into the derivative to find the marginal revenue when \( n = 100 \):\[R'(100) = 0.4 - 0.002 \times 100 = 0.4 - 0.2 = 0.2\]Thus, the marginal revenue when \( n = 100 \) is 0.2 dollars per computer.

Key concepts

Instantaneous Rate of Change

The concept of the instantaneous rate of change is crucial in understanding how functions behave. It is essentially the rate at which one quantity changes with respect to another. In simpler terms, it shows how fast or slow something is changing at a particular point.
In the context of revenue, this concept allows us to see how the revenue changes when we produce one more item. When you hear about the instantaneous rate of change in economic terms, it is often referred to as marginal concepts—like marginal revenue in this case. To find the instantaneous rate of change at a specific point, we use differentiation. This process helps reveal how a function's output changes with small changes in its input. By evaluating the derivative at a particular point, you can determine the instantaneous rate at that location, giving you crucial insights into the dynamics of your function at that exact moment.

Derivative of Functions

Differentiating a function involves finding another function, called the derivative, which tells us the rate of change of the original function. This rate of change is a central idea in calculus.
The derivative of a function provides a mathematical formula that shows how the function's value changes as its input changes. In our earlier solution, we are looking at a revenue function given by \( R(n) = 0.4n - 0.001n^2 \). To find the marginal revenue, which is the derivative of the revenue function with respect to production \( n \), the steps involve applying basic differentiation rules:

• The derivative of \( 0.4n \) is \( 0.4 \).
• The derivative of \(-0.001n^2 \) is \(-0.002n \).

So, the derivative \( R'(n) = 0.4 - 0.002n \) represents the marginal revenue function. By differentiating, you determine the precise rate at which revenue changes per unit produced, allowing for strategic economic decisions.

Economic Applications of Calculus

In economics, calculus is a powerful tool used to model and analyze different financial and production scenarios. Concepts like the derivative bring a deep understanding of marginal changes, which is fundamental for decision-making in business. For example:

• **Marginal Revenue**: This is the additional revenue that a company earns by selling one more unit of product. Calculating marginal revenue helps businesses understand the potential profitability of increasing production.
• **Marginal Cost**: Similarly, businesses use derivatives to determine the increase in cost incurred by producing one additional unit, which is the marginal cost. Understanding marginal cost alongside marginal revenue helps in finding the optimal level of production.

By utilizing calculus, economists and business analysts can precisely calculate key marginal values. These calculations support improved strategy, ensuring efficient resource allocation and maximizing profit. With these tools, firms can pinpoint when and how much to produce to achieve the best financial outcomes.