Problem 9 - Applied Calculus

Question

A company estimates that the total revenue, \(R\), in dollars. received from the sale of \(q\) items is \(R=\ln \left(1+1000 q^{2}\right)\). Calculate and interpret the marginal revenue if \(q=10\).

Step-by-Step Solution

Answer

The marginal revenue at \( q = 10 \) is approximately 0.2 dollars.

Step 1: Interpretation of Marginal Revenue

Marginal revenue is the derivative of the revenue function with respect to quantity, denoted as \( \frac{dR}{dq} \). It represents how much the revenue changes when the quantity sold increases by one unit.

Step 2: Revenue Function Definition

The total revenue function given is \( R = \ln(1 + 1000q^2) \). We need to find the derivative of this function to determine the marginal revenue.

Step 3: Differentiating the Revenue Function

To find the marginal revenue, compute the derivative of \( R \) with respect to \( q \). Let \( u = 1 + 1000q^2 \), then \( R = \ln(u) \). The derivative \( \frac{dR}{dq} \) is given by the chain rule: \( \frac{1}{u} \times \frac{du}{dq} \).

Step 4: Applying the Chain Rule

First, find \( \frac{du}{dq} = \frac{d}{dq}(1 + 1000q^2) = 2000q \). Then, using the chain rule, \( \frac{dR}{dq} = \frac{1}{1 + 1000q^2} \times 2000q \). Simplifying, we get \( \frac{dR}{dq} = \frac{2000q}{1 + 1000q^2} \).

Step 5: Substituting the Given Quantity

Substitute \( q = 10 \) into the marginal revenue formula: \( \frac{dR}{dq} = \frac{2000 imes 10}{1 + 1000 imes 10^2} = \frac{20000}{1 + 100000} = \frac{20000}{100001} \).

Step 6: Simplifying the Result

Divide the numerator by the denominator to approximate the marginal revenue: \( \frac{20000}{100001} \approx 0.2 \). This means that at \( q = 10 \), the marginal revenue is approximately 0.2 dollars.

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus used to find how a function changes as its input changes. It focuses on finding the derivative, which essentially measures the rate of change. When we talk about differentiating a function, we're concerned with finding this derivative.

In the context of our problem, we are interested in how the revenue function changes as the number of items sold changes. Mathematically, this is expressed as \( \frac{dR}{dq} \), where \( R \) is the revenue function and \( q \) is the quantity of items. By calculating this derivative, we determine the marginal revenue.

Marginal revenue tells us how much additional revenue is generated by selling one more unit. It helps businesses understand the dynamics of their revenue generation at different levels of production.

Revenue Function

The revenue function in business and economics describes how revenue varies with changes in the quantity of goods or services sold. In our exercise, the revenue function is given as \( R = \ln(1 + 1000q^2) \). This function incorporates a natural logarithm, indicating a nonlinear relationship between the revenue and the quantity sold.

This natural logarithm function shows that revenue increases quickly at small quantities but grows slowly at larger quantities, due to its log properties. This is crucial for understanding the real-world business implication: after a certain point, increased sales lead to diminishing revenue returns.

By using this revenue function in differentiation, we can analyze and predict the changes in revenue with each additional unit sold. Understanding your revenue function is like having a map; it guides decisions on pricing and sales strategy.

Chain Rule

The chain rule is a technique in calculus used to differentiate compositions of functions. It is vital when functions are nested within each other, just like in our exercise where \( u = 1 + 1000q^2 \) is inside the logarithm function.

To apply the chain rule, we break the differentiation task into parts. First, differentiate the outer function considering the inner function as a single variable. Then, differentiate the inner function itself.

In our exercise, the chain rule helps us find the derivative of \( R = \ln(u) \), where \( u = 1 + 1000q^2 \). We proceed by finding \( \frac{1}{u} \) from \( \ln(u) \), then multiply it by \( \frac{du}{dq} \), which is the derivative of the inner function. This simplification gives the marginal revenue formula: \( \frac{dR}{dq} = \frac{2000q}{1 + 1000q^2} \).

This technique is powerful and frequently used for solving complex real-world problems that involve nested functions.