Problem 56 - University Calculus: Early Transcendentals

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 average cost is 10 items.

Step 1: Calculate Average Cost

The average cost function, \( AC(x) \), is given by the total cost \( c(x) \) divided by the number of items \( x \). Thus, the formula for average cost is: \( AC(x) = \frac{c(x)}{x} = \frac{x^3 - 20x^2 + 20,000x}{x} = x^2 - 20x + 20,000 \).

Step 2: Differentiate the Average Cost Function

To find the production level that minimizes the average cost, we need to take the derivative of \( AC(x) \). The derivative \( AC'(x) \) is \( 2x - 20 \).

Step 3: Find Critical Points

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

Step 4: Determine the Nature of Critical Point

To determine if this critical point minimizes the average cost, check the second derivative, \( AC''(x) = 2 \). Since \( AC''(x) > 0 \), the function is concave up, indicating a minimum at \( x = 10 \).

Key Concepts

Cost Function

The cost function is a mathematical representation of the total cost associated with producing a certain number of items. In our exercise, the cost function is defined as \( c(x) = x^3 - 20x^2 + 20,000x \). This particular function is a polynomial, which is common in modeling real-world scenarios where variable costs increase with production levels.
Understanding this function is crucial because it lays the foundation for further analysis. The goal for this exercise is to find the production level that minimizes the average cost, thus making the whole manufacturing process more efficient. By analyzing and manipulating the cost function, we can derive insights into the cost behavior at different production levels. This insight helps businesses optimize their production to save on costs.

Derivative Calculus

Derivative calculus is employed here to find the rate of change of the average cost function. The derivative provides us with a tool to understand how the average cost changes as the production level \( x \) changes. In this context, the derivative of the average cost function, denoted by \( AC'(x) \), is calculated. It is found to be \( 2x - 20 \). The derivative tells us how the average cost behaves as the number of items produced changes, allowing us to pinpoint critical points where costs might be minimized.
Using derivatives is a standard technique to find optimal points in many problems, not just those concerning manufacturing costs. It's a versatile technique that provides a lot of insights into the problem being analyzed.

Critical Points

Critical points occur where the derivative of a function is zero or undefined. They are crucial in determining where a function might reach its minimum or maximum values. In our exercise, the critical point is found by setting \( AC'(x) = 2x - 20 = 0 \). Solving this equation gives us \( x = 10 \). These points are where the average cost isn't increasing or decreasing, which means they could represent a minimum, maximum, or an inflection point for the cost.
Identifying critical points is the first step towards understanding the overall optimization problem. It allows us to find potential points where the average cost might be minimized, which is the primary goal of this exercise.

Second Derivative Test

The second derivative test helps determine the nature of a critical point found in the previous steps. For our cost minimization problem, we find the second derivative \( AC''(x) \). In our example, the second derivative is \( 2 \), which is greater than zero. The test states that if the second derivative is positive at a critical point, then the function is concave upwards at that point, indicating a local minimum. Conversely, if it were negative, the function would be concave downwards, indicating a local maximum. In this case, since \( AC''(x) = 2 \), it confirms that the critical point at \( x = 10 \) is indeed a minimum. This validation is crucial as it concludes the optimization process by determining the optimal production level to minimize costs.