Problem 62 - Calculus and its applications

Question

Solid Rock Industries determines that the marginal cost, \(C^{\prime}\), of producing the xth climbing harness is given by $$C^{\prime}(x)=x^{3}-x$$ Find the total-cost function, \(C\), assuming that \(C(x)\) is in dollars and that fixed costs are \(\$ 6500\).

Step-by-Step Solution

Answer

The total cost function is \( C(x) = \frac{x^4}{4} - \frac{x^2}{2} + 6500 \).

Step 1: Understand Marginal Cost

The marginal cost function, denoted by \( C'(x) \), represents the derivative of the total cost function \( C(x) \) with respect to \( x \). It tells us the cost of producing one more unit.

Step 2: Set Up the Integration

Since \( C'(x) = x^3 - x \), to find the total cost function \( C(x) \), we will integrate the marginal cost function. This is because integration is the inverse operation of differentiation.

Step 3: Perform the Integration

Integrate the function \( C'(x) = x^3 - x \): \[C(x) = \int (x^3 - x) \, dx.\]This results in:\[C(x) = \frac{x^4}{4} - \frac{x^2}{2} + C_0,\]where \( C_0 \) is the constant of integration.

Step 4: Apply Initial Condition

We are given that the fixed costs are \\(6500, which means the total cost when no units are produced is \\)6500. Therefore, \( C(0) = 6500 \). Substitute into the total cost function:\[6500 = \frac{0^4}{4} - \frac{0^2}{2} + C_0 = C_0.\]Thus, \( C_0 = 6500 \).

Step 5: Write the Total Cost Function

Now substitute \( C_0 = 6500 \) back into the expression for \( C(x) \):\[C(x) = \frac{x^4}{4} - \frac{x^2}{2} + 6500.\]This represents the total cost function for producing \( x \) climbing harnesses.

Key Concepts

Marginal Cost

The concept of marginal cost is a fundamental idea in economics and plays a crucial role in Integral Calculus when dealing with cost functions. Marginal cost, denoted often as \( C'(x) \), indicates the additional cost incurred in the production of one more unit of a good or service. In mathematics, it is the derivative of the total cost function, \( C(x) \), with respect to the number of units produced, \( x \).
For example, if the marginal cost function is given as \( C'(x) = x^3 - x \), it illustrates how costs change with varying production levels. Understanding this derivative form helps businesses determine efficient production scales and pricing strategies.
Marginal cost functions are integral in deciding whether producing an additional unit will increase overall profitability. By examining \( C'(x) \), companies can avoid making products that result in losses or inefficiencies.

Total Cost Function

To determine the total cost function, or \( C(x) \), we utilize integration, a principal technique in Calculus. The total cost function gives a cumulative view of all production costs up to a certain unit count, \( x \). By integrating the marginal cost function, we convert the rate of cost into a total cost.
Starting with the marginal cost function \( C'(x) = x^3 - x \), we integrate it:

• \( C(x) = \int (x^3 - x) \, dx \)
• \( C(x) = \frac{x^4}{4} - \frac{x^2}{2} + C_0 \)

Here, \( C_0 \) stands as the constant of integration, which is crucial for defining the function fully.
The total cost function represents the full monetary commitment to manufacture \( x \) products, considering all variable and fixed costs. It is invaluable for budgeting and financial forecasting.

Fixed Costs

Fixed costs refer to expenses that do not change with the level of goods or services produced. They represent permanent expenditures which businesses incur regardless of their output levels, such as rent, salaries, and machinery costs. Here, fixed costs are given as \( \$6500 \), meaning when no units are manufactured, these costs are already established.
In the context of the total cost function, fixed costs play the role of the integration constant \( C_0 \). After integrating the marginal cost function \( C'(x) \), we need to know this fixed cost to determine the specific total cost function.
Knowing \( C(0) = 6500 \), we substitute into our integrated function:

• \( 6500 = \frac{0^4}{4} - \frac{0^2}{2} + C_0 \)
• \( C_0 = 6500 \)

This cost helps illustrate the baseline cost a company faces before producing any additional units, thus vital for accurate financial analysis and planning.