Problem 59 - Calculus Early Transcendentals: Pearson New International Edition

Question

The total cost of producing and selling \(n\) units of a certain commodity per week is \(C(n)=1000+n^{2} / 1200 .\) Find the average cost, \(C(n) / n\), of each unit and the marginal cost at a production level of 800 units per week.

Step-by-Step Solution

Answer

The average cost is \(\frac{1000}{n} + \frac{n}{1200}\); marginal cost at 800 is \(\frac{4}{3}\).

Step 1: Understand the Problem Statement

We are given a cost function \(C(n) = 1000 + \frac{n^2}{1200}\), representing the total cost of producing \(n\) units. We need to find the average cost per unit \(\frac{C(n)}{n}\) and the marginal cost when \(n = 800\).

Step 2: Write the Average Cost Function

The average cost \(AC(n)\) is the total cost divided by the number of units, expressed as \(AC(n) = \frac{C(n)}{n} = \frac{1000 + \frac{n^2}{1200}}{n}\). Simplifying gives: \[ AC(n) = \frac{1000}{n} + \frac{n}{1200}. \]

Step 3: Simplify the Average Cost Expression

By simplifying further, we obtain \(AC(n) = \frac{1000}{n} + \frac{n}{1200}\). This represents the average cost of producing a single unit.

Step 4: Derive the Marginal Cost Function

The marginal cost \(MC(n)\) is found by taking the derivative of the total cost function with respect to \(n\). Compute \(MC(n) = \frac{d}{dn}[1000 + \frac{n^2}{1200}] = \frac{d}{dn}[\frac{n^2}{1200}] = \frac{2n}{1200}\). Simplifying gives \(MC(n) = \frac{n}{600}\).

Step 5: Calculate Marginal Cost at 800 Units

Substitute \(n = 800\) into the marginal cost function: \(MC(800) = \frac{800}{600} = \frac{4}{3}\). The marginal cost at 800 units is \(\frac{4}{3}\) per unit.

Key Concepts

Average Cost Function

The average cost function helps us determine how much, on average, each unit of a product costs to produce. To find the average cost per unit, we divide the total cost function, denoted by \( C(n) \), by the number of units \( n \). In our exercise, the total cost function is given as \( C(n) = 1000 + \frac{n^2}{1200} \). From here, we calculate the average cost (AC) by using:

• \( AC(n) = \frac{C(n)}{n} \)
• Substitute the given function: \( AC(n) = \frac{1000 + \frac{n^2}{1200}}{n} \)

By simplifying, we break it into parts: \( \frac{1000}{n} + \frac{n}{1200} \). Each part of this expression represents a component of the average cost per unit. Understanding the average cost function is crucial for determining how efficiently a company is producing its units.

Marginal Cost

Marginal cost provides insight into how much the cost will increase if we produce one more unit. It is derived by calculating the derivative of the total cost function with respect to \( n \), the number of units. This helps businesses understand their cost structure more deeply. For our given exercise, the marginal cost function \( MC(n) \) is derived as follows:

• Start with the total cost function: \( \frac{n^2}{1200} \)
• Take the derivative with respect to \( n \): \( \frac{d}{dn}[\frac{n^2}{1200}] = \frac{2n}{1200} \)
• Simplify to get \( \frac{n}{600} \)

Thus, the marginal cost function \( MC(n) = \frac{n}{600} \). Knowing the marginal cost assists companies in making critical production decisions by showing the cost of producing an additional unit.

Derivative

A derivative is a fundamental concept in calculus that represents the rate of change of a function. In the context of cost functions, it allows us to determine how costs change as the number of units produced changes. Taking the derivative of the total cost function helps in finding the marginal cost.For our problem, we took the derivative of the cost function:

• \( C(n) = 1000 + \frac{n^2}{1200} \)
• We used the rule that the derivative of \( n^2 \) with respect to \( n \) is \( 2n \)
• So, \( \frac{d}{dn}[\frac{n^2}{1200}] = \frac{2n}{1200} \)

Derivatives tell us how a change in one quantity affects another and are essential in economics and business for optimizing costs and resource allocation.

Cost Function

A cost function encapsulates the total cost incurred by producing a certain number of units. It includes fixed costs, which are constant regardless of production levels, and variable costs, which change with production volume. In this exercise, our cost function is given by \( C(n) = 1000 + \frac{n^2}{1200} \), where:

• The first term, 1000, represents the fixed costs of production.
• The second term, \( \frac{n^2}{1200} \), captures the variable costs dependent on the number of units \( n \).

Understanding the cost function is critical for businesses to assess how costs evolve with production, enabling efficient pricing and budgeting strategies. The cost function serves as a cornerstone for calculating average and marginal costs, two key metrics in cost analysis.