Problem 17 - 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 $0.20; selling one more unit adds this to total revenue.

Step 1: Understand the Concept of Marginal Revenue

Marginal revenue is the additional revenue that can be gained from selling one more unit of a product. It is the derivative of the revenue function with respect to the quantity of items, denoted as \( \frac{dR}{dq} \).

Step 2: Differentiate the Revenue Function

Given the revenue function \( R = \ln(1 + 1000q^2) \), we need to find its derivative with respect to \( q \). Using the chain rule, the derivative, \( \frac{dR}{dq} \), is: \[ \frac{dR}{dq} = \frac{1}{1 + 1000q^2} \cdot 2000q. \]

Step 3: Substitute the Given Quantity

Now substitute \( q = 10 \) into the derivative to calculate the marginal revenue: \[ \frac{dR}{dq} \bigg|_{q=10} = \frac{1}{1 + 1000(10)^2} \cdot 2000(10) = \frac{20000}{1 + 100000} = \frac{20000}{100001}. \]

Step 4: Simplify and Interpret the Result

Calculate the result of the expression. The marginal revenue at \( q = 10 \) evaluates to approximately \( 0.199998 \). This indicates that when selling one additional unit at \( q = 10 \), the company generates approximately $0.20 more in revenue.

Key Concepts

Understanding the Revenue Function

The revenue function is the mathematical expression that defines how the revenue of a business depends on the quantity of goods sold. In our example, the revenue function is given by \( R = \ln(1 + 1000q^2) \). Here, \( R \) represents the revenue, and \( q \) denotes the quantity of items sold. This particular function is a natural logarithmic function, which means the revenue changes logarithmically with the square of the number of items, adjusted by a constant factor.Knowing the revenue function helps businesses predict their earnings based on different quantities of products sold. It provides a clear way to compute potential revenue figures depending on various sales levels. This information is crucial for strategic decision-making, such as setting prices or determining production levels.

Applying the Chain Rule

The chain rule is a fundamental tool for differentiation in calculus. It is used when differentiating composite functions, which are functions built from other functions, like our revenue function. The revenue function \( R = \ln(1 + 1000q^2) \) requires the chain rule as it comprises an outer function (the natural log) and an inner function \( (1 + 1000q^2) \).To apply the chain rule, you take the derivative of the outer function, evaluating it at the inner function. Then multiply it by the derivative of the inner function.- The derivative of the outer function \( \ln(x) \) is \( \frac{1}{x} \).- The derivative of the inner function \( 1 + 1000q^2 \) with respect to \( q \) is \( 2000q \).Putting these together using the chain rule gives us:\[ \frac{dR}{dq} = \frac{1}{1 + 1000q^2} \cdot 2000q. \]This is how we find the rate at which the revenue changes with respect to the quantity.

Differentiation in Calculus

Calculus differentiation involves finding the derivative of a function. The derivative represents how a function changes as its input changes. It's a key concept for understanding rates of change, like how revenue changes with sales in the exercise.Differentiation is applied here to determine the marginal revenue, which is the derivative of the revenue function with respect to quantity, \( q \). Calculating \( \frac{dR}{dq} \) allows us to understand precisely how much additional revenue is generated when one additional unit is sold.In the context of our exercise, after finding the derivative, we substitute \( q = 10 \) into the result to get the marginal revenue at that particular quantity. This gives us a tangible figure, enabling the company to assess the viability of producing and selling more units at that sales level.

Economic Interpretation of Marginal Revenue

Marginal revenue is an economic concept that refers to the additional income received from selling one more unit of a product. In simple terms, it indicates how much extra money you make when you increase sales by one unit.From our exercise, the marginal revenue when \( q = 10 \) was approximately \( 0.20 \). This is interpreted as: for each additional unit sold beyond 10 units, the company earns about $0.20 more. Knowing marginal revenue helps businesses decide how much more to produce or whether to adjust their prices, as it's crucial for profit maximization.In practical business scenarios, achieving a higher marginal revenue than the cost of producing an extra unit means the company would likely increase production. However, if the marginal revenue is less, it might reconsider or decrease output. Therefore, understanding and calculating marginal revenue is a powerful tool for decision-making in business strategy.