Problem 57 - Brief Applied Calculus

Question

A company's cost function is \(C(x)=\sqrt{4 x^{2}+900}\) dollars, where \(x\) is the number of units. Find the marginal cost function and evaluate it at \(x=20\).

Step-by-Step Solution

Answer

The marginal cost at \(x=20\) is 1.6 dollars.

Step 1: Define Marginal Cost Function

The marginal cost function is the derivative of the cost function, \(C(x)\), with respect to \(x\). It represents the cost of producing one more unit.

Step 2: Differentiate the Cost Function

To find the marginal cost function, differentiate \(C(x)=\sqrt{4x^2+900}\) using the chain rule. The derivative of \(\sqrt{u}\) is \(\frac{1}{2\sqrt{u}}\times u'\), where \(u=4x^2+900\).

Step 3: Apply the Chain Rule

First, compute the derivative of the inside function, \(u=4x^2+900\). The derivative is \(u' = 8x\). Now apply: \(C'(x) = \frac{1}{2\sqrt{4x^2+900}} \times 8x = \frac{8x}{2\sqrt{4x^2+900}}\).

Step 4: Simplify the Marginal Cost Function

Simplify the expression to get the marginal cost function: \(C'(x) = \frac{4x}{\sqrt{4x^2+900}}\).

Step 5: Evaluate the Marginal Cost at x=20

Substitute \(x=20\) into the marginal cost function: \(C'(20) = \frac{4(20)}{\sqrt{4(20)^2+900}}\). Calculate: \(C'(20) = \frac{80}{\sqrt{1600+900}} = \frac{80}{\sqrt{2500}} = \frac{80}{50} = 1.6\).

Key Concepts

Cost Function

A cost function represents the total cost incurred by a company to produce a specific number of goods or services. In our scenario, the cost function is given as \(C(x)=\sqrt{4x^{2}+900}\), where \(x\) represents the number of units produced. Breaking down the function:

• \(4x^2\) refers to a variable cost element that depends on the number of units.
• The constant \(900\) could represent the fixed costs, which are the costs that stand regardless of the unit production.

Understanding a cost function is vital as it helps in calculating marginal costs, profit analysis, and decision-making regarding production levels.
The function helps businesses to ascertain how costs will behave with changes in production levels.

Derivative

A derivative in calculus represents the rate at which a function is changing at any given point, making it a crucial tool in understanding cost functions. Here, the derivative helps us find the marginal cost, which shows the additional cost incurred by producing one more unit.

• To find the derivative of a cost function like \(C(x)=\sqrt{4x^2+900}\), we need to apply differentiation rules.
• Differentiation provides insights into how costs will change with slight alterations in production, offering a powerful decision-making tool for businesses.

The derivative gives us the marginal cost function, a major component in cost analysis.

Chain Rule

The chain rule is a fundamental technique in calculus used to differentiate composite functions. It is especially useful in our scenario where the cost function \(C(x)=\sqrt{4x^2+900}\) involves a function inside another function.
The chain rule follows a straightforward process:

• Identify the inner function \(u = 4x^2 + 900\).
• Differentiating this gives us \(u' = 8x\).
• Then apply: \(C'(x) = \frac{1}{2\sqrt{u}} \times u'\).

This rule helps break down complex differentiation tasks into simpler parts. Understanding how to implement the chain rule is crucial for tackling various calculus problems.

Evaluate Function

The final step in many calculus problems involves evaluating the function at a certain point, providing specific values that are often critically needed for real-world applications. In this example, we evaluate the marginal cost function at \(x=20\).
To evaluate, simply substitute \(x\) with 20 in the marginal cost function. So, \(C'(20) = \frac{4(20)}{\sqrt{4(20)^2+900}}\).

• Calculate inside the square root: \(4(20)^2 + 900 = 2500\).
• Find the square root: \(\sqrt{2500} = 50\).
• Thus, \(C'(20) = \frac{80}{50} = 1.6\).

This calculation demonstrates the importance of evaluating for precise, interpretable results that guide managerial decisions and strategies.