Question

Marginal Revenue Suppose the weekly revenue in dollars from selling x custom- made office desks is \(r(x)=2000\left(1-\frac{1}{x+1}\right)\) (a) Draw the graph of \(r .\) What values of \(x\) make sense in this problem situation? (b) Find the marginal revenue when \(x\) desks are sold. (c) Use the function \(r^{\prime}(x)\) to estimate the increase in revenue that will result from increasing sales from 5 desks a week to 6 desks a week. (d) Writing to Learn Find the limit of \(r^{\prime}(x)\) as \(x \rightarrow \infty\) How would you interpret this number?

Short answer

The marginal revenue when \(x\) desks are sold is \(\frac{2000}{(x+1)^2}\). The addition of one more desk to the current number of 5 desks sold per week will add about $55.56 to the weekly revenue--this is the marginal revenue. Furthermore, as the number of desks sold increases without bound, the marginal revenue approaches zero. This suggests that the benefit in terms of revenue from selling an additional desk diminishes as the quantity of desks sold becomes large.

Step-by-step solution

Graph the revenue function and identify feasible values for \(x\)

The graph of the function \(r(x)=2000\left(1-\frac{1}{x+1}\right)\) is a decreasing curve. As \(x\) represents the number of desks sold, only non-negative values of \(x\) are valid. Specifically, \(x\) can range from 0 (no desks are sold) to some large positive number (there is no known upper limit for the number of desks that could be sold).

Find the marginal revenue

The marginal revenue is found by determining the derivative of the revenue function. So, \(r'(x) = \frac{d}{dx}\left[2000\left(1-\frac{1}{x+1}\right)\right] = \frac{2000}{(x+1)^2}\). This is the rate of change in revenue for each additional desk sold.

Estimate increase in revenue as sales increase by 1

To estimate the increase in revenue as sales increase from 5 desks to 6 desks, plug \(x=5\) into \(r'(x) = \frac{2000}{(x+1)^2}\) and compute. Hence, \(\Delta r = r'(5) = \frac{2000}{(5+1)^2} = \frac{2000}{36} \approx 55.56\) dollars. This suggests that an additional desk sold above 5 will bring in approximately $55.56 in revenue.

Find limit of marginal revenue as \(x \rightarrow \infty\)

The limit is given by \(\lim_{x\to\infty} r'(x) = \lim_{x\to\infty} \frac{2000}{(x+1)^2} = 0\). The interpretation of this result is that as the quantity of desks sold becomes very large, the increase in revenue for each additional desk sold tends to zero.

Key concepts

Revenue Function Graphing

Understanding the visual representation of revenue functions is fundamental for grasping how a business's income varies with sales. Graphing the revenue function, such as the provided example of selling custom-made office desks, helps identify the relationship between the quantity sold and the revenue generated.

Using our function, \(r(x)=2000\left(1-\frac{1}{x+1}\right)\), we can plot this as a curve on a graph, where the horizontal axis (x-axis) represents the number of desks sold, and the vertical axis (y-axis) represents the revenue in dollars. For this problem, only non-negative values of \(x\) are reasonable, since a negative number of desks cannot be sold. As more desks are sold, the revenue increases but at a decreasing rate. This tells us there are diminishing returns on the revenue made from each additional desk sold as the quantity increases.

Marginal Revenue Derivation

Marginal revenue is the additional income obtained from selling one more unit of a good or service. It is a critical concept in business and economics, representing the change in total revenue as a result of changing the quantity of the product sold by one unit.

To derive the marginal revenue, we differentiate the revenue function \(r(x)\) with respect to \(x\). For our example, the marginal revenue function \(r'(x)\) is computed as \(r'(x) = \frac{d}{dx}\left[2000\left(1-\frac{1}{x+1}\right)\right] = \frac{2000}{(x+1)^2}\). This formula allows you to find the rate at which revenue changes for each additional desk sold. The process of finding this derivative is a direct application of calculus, particularly the rules of differentiation.

Rate of Change in Revenue

The rate of change in revenue, often referred to as marginal revenue, is a pivotal concept for understanding how revenue evolves when sales quantities change. It is, essentially, the slope of the revenue function at any given point.

The formula derived for marginal revenue, \(r'(x) = \frac{2000}{(x+1)^2}\), tells us precisely how much the revenue is expected to increase with the sale of each additional desk. For example, when we plug in \(x=5\), we find that selling the sixth desk would increase revenue by approximately $55.56. This incremental view of revenue can help businesses make informed decisions about pricing and production levels.

Limit of a Function

In calculus, the concept of a limit explores the behavior of a function as the input approaches a certain value. Limits are foundational to all of calculus, from defining derivatives to understanding continuous functions.

In the context of our revenue problem, the limit of marginal revenue as \(x \rightarrow \infty\) is of interest. This limit, \(\lim_{x\to\infty} r'(x)\), helps us understand what happens to the additional revenue as the quantity sold increases without bound. In the case of the office desks, the limit is zero, meaning that as an extraordinarily large number of desks are sold, the revenue gained from selling one more desk is negligible. This reflection on the limit tells us about the long-term sustainability and pricing strategy that the business might need to consider.