Question

Find the change in cost \(C\), revenue \(R\), or profit \(P\), for the given marginal. In each case, assume that the number of units \(x\) increases by 3 from the specified value of \(x\). $$ \frac{d R}{d x}=75\left(20+\frac{900}{x}\right) \quad x=500 $$

Short answer

The change in revenue for an increase in three units can be found by calculating \( \Delta R = R(x+3) - R(x) = 75(20(x+3)+900\ln|x+3|) - 75(20x+900\ln|x|)\). Replacing \(x\) with the given value and simplifying the expression will give the desired result.

Step-by-step solution

Understand the Problem

We are given the marginal revenue function \( \frac{d R}{d x}=75(20+\frac{900}{x}) \) when \( x=500 \). This tells us how much the revenue increases for each additional unit sold. We are also told that the number of units \(x\) increases by 3 units, and we are asked to find the change in revenue.

Integration

We would integrate the marginal revenue equation to get the revenue function. This is done because the integral of the rate of change (the derivative) gives us the original function. The integral of \( \frac{d R}{d x}=75(20+\frac{900}{x}) \) is \( R = 75 \int (20+\frac{900}{x}) dx \). Performing the integration yields \( R = 75(20x+900\ln|x|) + C\) where \(C\) is the constant of integration.

Apply the Increase in Units

We then need to calculate the change in revenue when the number of units \(x\) increases by 3 from 500. This means we substitute \(x+3\) and \(x\) into the revenue equation and find the difference. The change in revenue \(\Delta R\) is thus \( \Delta R = R(x+3) - R(x) = 75(20(x+3)+900\ln|x+3|) - 75(20x+900\ln|x|)\). This value will give us the change in revenue for an increase of three units.

Key concepts

Marginal Analysis

Marginal analysis is a pivotal concept in economics focusing on the additional changes that occur with every incremental unit of a good or service. This tool is employed to determine the optimal level of production or consumption where the marginal benefit equals the marginal cost. When applied to revenue, marginal analysis looks at the extra revenue generated by selling one more unit, which is mathematically represented as the derivative of the revenue function with respect to the quantity, or \( \frac{dR}{dx} \). In the given exercise, the marginal revenue function is \( \frac{dR}{dx} = 75(20 + \frac{900}{x}) \), indicating that the revenue increase for each additional unit sold changes with different values of \(x\). This concept is crucial for businesses to understand how much they would expect to earn from increasing their sales by one unit at any point.

Through marginal analysis, companies can make informed decisions on whether the cost of producing one more item is justified by the additional revenue it generates, ultimately informing pricing strategies, production levels, and profit maximization efforts.

Revenue Function Integration

Integration in the context of a revenue function is a fundamental application of calculus to economics. When you're given a marginal revenue function, it can be integrated to find the general revenue function. This integration process essentially reverses differentiation, summing up the marginal revenues over all units to provide a formula for total revenue.

In the exercise, the provided marginal revenue function is integrated resulting in the formula \( R = 75(20x + 900\ln|x|) + C \), where \( C \) is the constant of integration that would be determined by initial conditions or known revenue at a specific quantity sold. The integration process here aids in establishing a broader understanding of how revenue accumulates across different quantities, providing a more comprehensive view of revenue beyond just the incremental changes.

Change in Revenue

The 'change in revenue' refers to the difference in total revenue that occurs as a result of selling additional units of a product or service. It is the actual additional income generated by increasing the sales volume. In practical applications, understanding this change helps in decision-making related to production and sales strategies.

The exercise demonstrates this concept by using the derived revenue function to calculate the change in revenue as a result of increasing the quantity by 3 units. By evaluating the revenue function at \( x+3 \) and \( x \) and finding the difference \( \Delta R \), we get the actual increase in total revenue associated with a small increase in quantity, assuming all other factors remain constant. This is a direct application of the concept in a real-world context, illustrating the financial impact of production decisions.

Differential Calculus in Economics

Differential calculus is a branch of mathematics that deals with the rates at which quantities change. In economics, it is used extensively to analyze the behavior of economic variables, such as cost, revenue, and profit, in response to changes in other variables like price or output quantity.

The marginal revenue function in the given exercise is a manifestation of differential calculus, as it describes the rate of change of revenue with respect to the number of units sold. Derivatives provide crucial insights into the responsiveness of economic factors, enabling economists and businesses to predict and respond to market dynamics effectively. This application of calculus is instrumental for optimizing economic outcomes and strategic planning.