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 aver cost is minimized.

Step-by-Step Solution

Answer

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

Step 1: Understanding the Function

The function given is the total cost of production in thousands, represented by \( C(q) = q^3 - 12q^2 + 60q \). Here, \( q \) is the quantity in thousands and ranges from 0 to 8. Our task is to analyze this function to find the quantity \( q \) where the average cost is minimized.

Step 2: Graphing the Function

To graph the function \( C(q) \), plot the graph for \( q \) values ranging from 0 to 8. Plot points for \( C(q) \) by substituting \( q \) into the equation, and connect these points to form the curve. Observing the graph will allow us to visually estimate the value of \( q \) where average cost seems to be minimized.

Step 3: Calculating the Average Cost Function

The average cost function \( AC(q) \) is calculated as the total cost divided by quantity, \( AC(q) = \frac{C(q)}{q} = \frac{q^3 - 12q^2 + 60q}{q} \). Simplifying, we get \( AC(q) = q^2 - 12q + 60 \) for \( q eq 0 \).

Step 4: Minimizing the Average Cost Analytically

To find the \( q \) that minimizes \( AC(q) = q^2 - 12q + 60 \), take the derivative \( AC'(q) = 2q - 12 \), and set it to zero for critical points: \( 2q - 12 = 0 \). Solving gives \( q = 6 \). Check the second derivative \( AC''(q) = 2 \), which is positive, indicating a local minimum at \( q = 6 \).

Step 5: Confirmation by Graph and Analysis

Confirm by comparing this analytical solution with the graph from Step 2. The graph should show a dip in the average cost at around \( q = 6 \), confirming the correctness of the analytical solution.

Key Concepts

Cost Function

In applied calculus, a cost function is a mathematical representation of how costs change depending on the production level. For our problem, the cost function is given by \( C(q) = q^3 - 12q^2 + 60q \). It encapsulates the total cost spent in thousands of dollars when producing \( q \) units, where \( q \) is also measured in thousands and is bounded between 0 and 8.
Understanding this function is pivotal because it allows businesses to predict expenses associated with varying production levels. It consists of multiple terms:

• \( q^3 \) term signifies how cost increases with higher production, likely due to inefficiencies or limitations arising at scale.
• \(- 12q^2 \) may represent decreasing returns to scale or cost reductions from economies of scale until certain constraints kick in.
• \( + 60q \) can be linked to the variable costs, directly proportional to production.

Analyzing such a function forms the foundation for more complex economic and business decisions.

Graph Analysis

Graphing the cost function \( C(q) \) across values from 0 to 8 gives a visual interpretation of how costs behave as production changes. To create the graph, substitute \( q \) into the equation and determine points for various \( q \) values. Connect these points to form a curve.
The graph provides intuitive insights:

• A visually identifiable curve showing where costs increase or decrease with production.
• The point where the graph shows the least cost per unit signifies the quantity where average cost is minimized.

In practice, graph analysis complements mathematical calculations, providing businesses with a clearer picture of trends, aiding in more tangible decision-making processes. Visual estimations often suggest initial hypotheses that are further confirmed through analytical methods.

Derivative

A derivative in calculus expresses how a function's value changes as its input changes. For applied calc, it involves understanding rates of change within cost functions. To minimize the average cost function, one needs to differentiate it.In our problem, after simplifying the average cost function to \( AC(q) = q^2 - 12q + 60 \), we differentiate with respect to \( q \) to obtain \( AC'(q) = 2q - 12 \).
Finding the derivative helps in identifying critical points, where the rate of change is zero, indicating a potential maximum or minimum.

This critical point is crucial for determining the efficiencies in production levels:

• Setting \( AC'(q) = 0 \) yields critical points.
• In this scenario, \( q = 6 \) is obtained.
• The second derivative \( AC''(q) = 2 \) being positive confirms it's a minimum.

Mastery of derivatives allows students to navigate cost optimization problems systematically, deeply enhancing analytical capabilities.

Average Cost Minimization

Minimizing average cost means identifying the optimal production level where the cost per unit is least. It's a key concept in production economics, allowing companies to produce efficiently.The average cost function derived from the total cost \( C(q) \) is \( AC(q) = \frac{q^3 - 12q^2 + 60q}{q} = q^2 - 12q + 60 \).
The task is to find the \( q \) which minimizes this function. Here's the step-by-step process:

• Calculate the derivative of \( AC(q) \).
• Solve \( AC'(q) = 2q - 12 = 0 \) to get \( q = 6 \).
• Verify with the second derivative \( AC''(q) = 2 \), which shows a minimum at \( q = 6 \).

Replacing analytical with visual, the graph also displays this efficiency at \( q = 6 \), aligning with the calculated result. Mastering such analyses empowers a sound decision-making framework, optimal for academic and real-world applications.