Problem 20 - Calculus Early Transcendentals: Pearson New International Edition

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.)

Step-by-Step Solution

Answer

Marginal revenue: 0.38 dollars per computer at \(n=10\), 0.2 dollars per computer at \(n=100\).

Step 1: Understand the Revenue Function

The given revenue function is \(R(n) = 0.4n - 0.001n^2\), where \(n\) represents the number of computers produced. We need to find the derivative \(R'(n)\) to determine the marginal revenue, which is the instantaneous rate of change of revenue with respect to \(n\).

Step 2: Differentiate the Revenue Function

To find \(R'(n)\), take the derivative of \(R(n) = 0.4n - 0.001n^2\). Using basic differentiation rules: for \(0.4n\), the derivative is \(0.4\), and for \(-0.001n^2\), using the power rule, the derivative is \(-0.001 \times 2n = -0.002n\). Thus, \(R'(n) = 0.4 - 0.002n\).

Step 3: Calculate Marginal Revenue at n=10

Substitute \(n = 10\) into the derivative to find the marginal revenue at this production level: \(R'(10) = 0.4 - 0.002 \times 10 = 0.4 - 0.02 = 0.38\). Thus, the marginal revenue when \(n = 10\) is \(0.38\) dollars per computer.

Step 4: Calculate Marginal Revenue at n=100

Substitute \(n = 100\) into the derivative to find the marginal revenue at this production level: \(R'(100) = 0.4 - 0.002 \times 100 = 0.4 - 0.2 = 0.2\). Therefore, the marginal revenue when \(n = 100\) is \(0.2\) dollars per computer.

Key Concepts

Instantaneous Rate of Change

The instantaneous rate of change is a crucial concept when analyzing how a quantity changes at a specific point. Imagine you're riding a bike on a winding road. Your speedometer tells you the speed at that very moment - that's a type of an instantaneous measurement. Similarly, in business, the instantaneous rate of change helps you understand how quickly revenue grows or declines as production changes.
The idea is to get a snapshot of change at an exact moment rather than over a longer period. This is particularly helpful when making quick decisions about production levels. In the context of our problem, the rate of change in revenue is called the marginal revenue.
With a mathematical approach, this rate is captured by finding the derivative of the revenue function. This derivative gives us a function that tells us the rate of change of revenue for every single unit increase in production.

Differentiation

Differentiation is a process in calculus that involves finding the derivative of a function. It's like pulling back a curtain to see what is happening to the output of a function as its input changes.
Consider the revenue function in our exercise, given by: \[ R(n) = 0.4n - 0.001n^2 \]To find the marginal revenue, we differentiate this function with respect to \(n\). Differentiation here uses two simple rules:

• For \(0.4n\), a linear term, the derivative is simply the coefficient, which is \(0.4\).
• For \(-0.001n^2\), the power rule of differentiation applies. This rule states that the derivative of \(ax^b\) is \(abx^{b-1}\), resulting in \(-0.002n\) for our term.

By applying these rules, the derivative, or \(R'(n)\), is formed as:\[ R'(n) = 0.4 - 0.002n \]This expression allows us to calculate the marginal revenue at any point \(n\) by substituting \(n\) into this derivative equation.

Revenue Function

The revenue function in an economic context represents how much money a company makes from selling its products. Our given example is \[ R(n) = 0.4n - 0.001n^2 \],where \(R(n)\) is the revenue and \(n\) is the number of computers produced.
Every term in this function tells a part of a story:

• The first term, \(0.4n\), implies that each computer contributes \(0.4\) dollars to the revenue, assuming all are sold.
• The second term, \(-0.001n^2\), represents a diminishing return effect, where producing more units adds a non-linear negative impact on revenue. This could be due to rising costs or saturation in the market.

Understanding the revenue function helps in decision-making by showing the expected income for different production levels. By knowing both the function and its derivative, businesses can effectively strategize how to maintain or improve profitability by choosing optimal production levels.