Problem 60 - 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 Average Cost Function

The average cost function is given by \( AC(x) = \frac{c(x)}{x} \). For the given cost function \( c(x) = x^{3} - 20x^{2} + 20000x \), the average cost function becomes:\[AC(x) = \frac{x^{3} - 20x^{2} + 20000x}{x} = x^{2} - 20x + 20000.\]

Step 2: Find the Derivative of Average Cost

To find the minimum average cost, we need to find the derivative of the average cost function, \( AC'(x) \). Calculate:\[AC'(x) = \frac{d}{dx} (x^{2} - 20x + 20000) = 2x - 20.\]

Step 3: Set the Derivative to Zero

Set the derivative \( AC'(x) = 2x - 20 \) to zero to find the critical points that potentially minimize the average cost:\[2x - 20 = 0.\]

Step 4: Solve for the Critical Points

Solve the equation \(2x - 20 = 0\) to find the value of \(x\):\[2x = 20 \Rightarrow x = 10.\]

Step 5: Evaluate the Nature of the Critical Point

To determine whether \(x = 10\) minimizes the average cost, evaluate the second derivative \(AC''(x)\):\[AC''(x) = 2.\]Since \(AC''(x) > 0\), the function is concave up at \(x = 10\), indicating a local minimum.

Key Concepts

Cost Function

The cost function, denoted as \( c(x) \), is crucial in understanding the cost structure of producing items. It represents the total cost incurred when manufacturing \( x \) items. In this exercise, the cost function is given by \( c(x) = x^3 - 20x^2 + 20000x \).
To analyze the cost function, consider each term:

• The term \( x^3 \) might suggest variable costs that increase disproportionately with each additional unit.
• The \(-20x^2\) term might indicate a reduction in costs possible through economies of scale.
• The \( +20000x \) term often represents linear, direct costs.

Cost functions are used to calculate the average cost, which is the cost per unit of production. This is essential for businesses to determine the efficiency and the profitability of manufacturing at different levels.

Critical Points

Critical points play a pivotal role in optimization problems such as average cost minimization. These are the points where the derivative of a function is zero or undefined, indicating the potential for maximum, minimum, or saddle points.
In our exercise, we find the critical points by setting the derivative of the average cost function, \( AC'(x) = 2x - 20 \), equal to zero:

• \( 2x - 20 = 0 \) results in \( x = 10 \).

This critical point \( x = 10 \) is where we suspect a minimum average cost. However, further analysis via the second derivative is necessary to confirm the nature of this critical point.
Understanding critical points helps identify where a function's behavior changes, marking the transition between increasing and decreasing intervals.

Derivative

The derivative of a function is a fundamental concept in calculus, used to determine the rate of change. In the context of this problem, the derivative is key to finding the points where the average cost is minimized.
Given the average cost function \( AC(x) = x^2 - 20x + 20000 \), its derivative, \( AC'(x) \), is computed as \( 2x - 20 \).
Why is this important? The derivative tells us how the average cost changes as the production level \( x \) changes:

• If the derivative is positive, the average cost increases with production.
• If negative, the cost decreases.

By setting the derivative to zero, one identifies critical points, representing potential minima or maxima.
The derivative provides insights into the behavior of cost functions, enabling the determination of efficient production levels.

Second Derivative Test

The second derivative test helps confirm the nature of critical points determined from the first derivative. Specifically, it indicates whether a point is a minimum, maximum, or saddle point.
To apply this test, we calculate the second derivative of the average cost function. Here, \( AC''(x) = 2 \) is the second derivative, which is constant and greater than zero.
What does this imply?

• Since the second derivative is positive, the function is concave up at \( x = 10 \), indicating a relative minimum.

This confirmation means that at \( x = 10 \), the average cost of production is minimized.
Utilizing the second derivative test is a robust method to verify the nature of critical points and ensure accurate optimization conclusions in cost analysis scenarios.