Problem 48 - Thomas Calculus

Question

Production level Suppose that \(c(x)=x^{3}-20 x^{2}+20,000 x\) is the cost of manufacturing \(x\) items. Find a production level that will minimize the average cost of making \(x\) items.

Step-by-Step Solution

Answer

The production level that minimizes the average cost is 10 items.

Step 1: Define the Average Cost Function

The average cost function is given by the total cost divided by the number of items produced. Thus, if the cost function is denoted as \(c(x) = x^3 - 20x^2 + 20000x\), the average cost function \(A(x)\) is calculated as follows:\[ A(x) = \frac{c(x)}{x} = \frac{x^3 - 20x^2 + 20000x}{x} = x^2 - 20x + 20000.\]

Step 2: Determine the Derivative of the Average Cost Function

To find the minimum of the average cost function, we first need to take its derivative with respect to \(x\). The derivative of \(A(x) = x^2 - 20x + 20000\) is:\[ A'(x) = 2x - 20.\]

Step 3: Solve for Critical Points

Set the derivative \(A'(x)\) equal to zero to find critical points:\[ 2x - 20 = 0.\]Solve for \(x\):\[ 2x = 20 \quad \Rightarrow \quad x = 10.\]

Step 4: Verify Minimum with Second Derivative Test

To ensure that this critical point corresponds to a minimum, we check the second derivative, \(A''(x)\) of the average cost function:\[ A''(x) = 2.\]Since \(A''(x) > 0\), the average cost function has a minimum at \(x = 10\).

Step 5: Interpret the Results

The production level of \(x = 10\) items minimizes the average cost of manufacturing. This means the average cost per item is at its lowest when 10 items are produced.

Key Concepts

Average Cost Function

The average cost function is a vital concept in understanding how costs behave over different levels of production. It helps businesses determine how much, on average, each item costs to produce. Simply put, the average cost is the total cost divided by the number of items produced. In the given exercise, the cost function is provided as \(c(x) = x^3 - 20x^2 + 20000x\). To find the average cost function, represented as \(A(x)\), we divide the total cost \(c(x)\) by \(x\). This simplifies to:

• \(A(x) = \frac{c(x)}{x} = \frac{x^3 - 20x^2 + 20000x}{x} = x^2 - 20x + 20000\).

Understanding this function allows us to explore the behavior of costs as production increases, guiding decisions for optimal production levels.

Derivative

In calculus, a derivative represents how a function's output changes as the input changes. When optimizing a function like the average cost function, finding its derivative is crucial.
The derivative, denoted as \(A'(x)\), helps us identify points where the function might reach its minimum or maximum. In our exercise, the average cost function \(A(x) = x^2 - 20x + 20000\) has a derivative:

• \(A'(x) = 2x - 20\).

This expression tells us how each increase in \(x\) affects the average cost. By calculating this derivative, we aim to find out which production level minimizes costs. That is why finding the derivative is the first step towards optimization.

Critical Points

Critical points are where the derivative of a function equals zero or is undefined, indicating potential minima or maxima. These points are essential when determining optimal conditions like minimizing average cost. For our example, we set \(A'(x) = 2x - 20\) equal to zero to find these critical points:

• \(2x - 20 = 0\).

Solving this equation gives us \(x = 10\).
This means that producing 10 items may minimize the average cost. Critical points guide us in pinpointing where to check for the actual minimum or maximum of the cost function.

Second Derivative Test

The second derivative test is a method used to determine whether a critical point is a minimum, maximum, or neither. This step involves taking the derivative of the derivative, known as the second derivative. When applied to the average cost function, if the second derivative is positive at a critical point, the function has a local minimum.
For our exercise, the second derivative of \(A(x)\) is:

• \(A''(x) = 2\).

Since \(A''(x) = 2\) is greater than zero, it confirms that the critical point \(x = 10\) is indeed a minimum. This means producing 10 items minimizes the average cost, verifying our previous calculations. This test is a reliable way to conclude that we have found the optimal production level.