Problem 7 - Applied Calculus

Question

The total cost of production, in thousands of dollars, is \(C(q)=q^{3}-12 q^{2}+60 q\), where \(q\) is in thousands and \(0 \leq q \leq 8\) (a) Graph \(C(q)\). Estimate visually the quantity at which average cost is minimized. (b) Determine analytically the exact value of \(q\) at which average cost is minimized.

Step-by-Step Solution

Answer

The average cost is minimized when \(q = 6\).

Step 1: Understanding the Problem

We are given the cost function for production \(C(q) = q^3 - 12q^2 + 60q\). The task is to first estimate and then determine the quantity \(q\) that minimizes the average cost.

Step 2: Estimate from Graph

Plot the function \(C(q) = q^3 - 12q^2 + 60q\) for \(0 \leq q \leq 8\). Observe the curve to identify a value of \(q\) where the average cost appears to be minimized. Let's assume that graphing gives an approximate value (such as around \(q = 4\)).

Step 3: Define the Average Cost

The average cost \(AC(q)\) is defined as \(AC(q) = \frac{C(q)}{q}\). Simplifying this formula using the given \(C(q)\), we have:\[ AC(q) = \frac{q^3 - 12q^2 + 60q}{q} = q^2 - 12q + 60 \] for \(q > 0\).

Step 4: Differentiate Average Cost

Find the derivative of \(AC(q)\), \(AC'(q)\), to find critical points. This involves differentiating:\[ AC'(q) = \frac{d}{dq}(q^2 - 12q + 60) = 2q - 12 \]

Step 5: Solve for Critical Points

Set the derivative equal to zero to find critical points:\[ 2q - 12 = 0 \]\[ 2q = 12 \]\[ q = 6 \]

Step 6: Verify Minimum with Second Derivative

Use the second derivative test to ensure this point is a minimum. The second derivative:\[ AC''(q) = \frac{d^2}{dq^2}(q^2 - 12q + 60) = 2 \]Since \(AC''(q) = 2 > 0\), \(AC(q)\) is minimized at \(q = 6\).

Step 7: Conclusion and Comparison

From our graph we estimated \(q\) around 4 might minimize costs, but analytically, we found that the exact minimum average cost occurs at \(q = 6\).

Key Concepts

Cost Function

A cost function is a mathematical formula that businesses use to predict and analyze the costs associated with producing a set of goods. In our given exercise, the cost function is given by: \( C(q) = q^3 - 12q^2 + 60q \), where \(q\) represents the quantity produced in thousands. **Understanding this function is essential.**

• The terms \( q^3 \), \( -12q^2 \), and \( 60q \) are components that define how costs change with respect to quantity.
• The cube term \( q^3 \) often suggests increasing costs at higher production levels.
• The term \( -12q^2 \) reflects a decrease in costs initially, due to factors like economies of scale.
• The linear term \( 60q \) is usually associated with variable costs that increase with production.

Understanding how these parts work together helps in anticipating how the total cost behaves with varying production levels. The visual representation of \( C(q) \) aids in estimating points of interest, such as where average costs might be minimized.

Critical Points

Critical points are values of \( q \) where the rate of change of the function, the derivative, is zero or undefined. **These points are significant**, as they can indicate potential minimum or maximum values of the function. To find critical points in our exercise, we differentiate the average cost function \( AC(q) \). After simplifying the average cost function to \( AC(q) = q^2 - 12q + 60 \), the derivative is calculated: \[ AC'(q) = 2q - 12 \] Setting this derivative equal to zero determines the critical points:

• \( 2q - 12 = 0 \)
• Solving gives \( q = 6 \)

The critical point at \( q = 6 \) needs further verification to confirm whether it is indeed a minimum. This is where the second derivative test comes into play.

Second Derivative Test

The second derivative test is a mathematical method used to confirm the nature of critical points. **By examining the concavity** of the function at these points, we conclude whether they are minima or maxima. In our exercise, the average cost function's first derivative \( AC'(q) \) is \( 2q - 12 \). Taking the second derivative yields: \[ AC''(q) = 2 \]

• If \( AC''(q) > 0 \) at the critical point, the function has a local minimum.
• If \( AC''(q) < 0 \) at the critical point, the function has a local maximum.
• If \( AC''(q) = 0 \), the test is inconclusive.

Here, since \( AC''(q) = 2 \) and is greater than zero, we confirm that the average cost function has a local minimum at \( q = 6 \). This substantiates our analytical result that the average cost is minimized at this point.

Average Cost Function

The average cost function is designed to measure the cost per unit of production. It provides insights into how efficiently resources are being used in the production process. In essence, the average cost at any production level \( q \) is calculated as: \[ AC(q) = \frac{C(q)}{q} \] In our specific exercise, substituting the given \( C(q) \) into the formula yields: \[ AC(q) = \frac{q^3 - 12q^2 + 60q}{q} = q^2 - 12q + 60 \]

• This function reveals how the average cost varies as production changes.
• It highlights the relationship between total production cost and the quantity produced.
• The simplified function \( q^2 - 12q + 60 \) assists in calculating derivatives and analyzing cost mitigation strategies.

Analyzing the average cost function is crucial for businesses intending to optimize production costs, as it pinpoints the production level at which the cost per unit is minimized.