Problem 9 - Applied Calculus

Question

The average cost per item to produce \(q\) items is given by $$ a(q)=0.01 q^{2}-0.6 q+13, \text { for } q>0 $$ (a) What is the total cost, \(C(q)\), of producing \(q\) goods? (b) What is the minimum marginal cost? What is the practical interpretation of this result? (c) At what production level is the average cost a minimum? What is the lowest average cost? (d) Compute the marginal cost at \(q=30 .\) How does this relate to your answer to part (c)? Explain this relationship both analytically and in words.

Step-by-Step Solution

Answer

(a) The total cost is \(C(q) = 0.01q^3 - 0.6q^2 + 13q\). (b) Minimum marginal cost is at \(q=20\). It shows optimal economies of scale. (c) The lowest average cost is 4 at \(q=30\). (d) Marginal cost at \(q=30\) is 4, indicating efficient production.

Step 1: Understanding Average Cost Function

The average cost function is given by \(a(q) = 0.01q^2 - 0.6q + 13\). This function represents the cost per unit of producing \(q\) items.

Step 2: Finding Total Cost, C(q)

To find the total cost \(C(q)\), multiply the average cost \(a(q)\) by \(q\): \[C(q) = q(0.01q^2 - 0.6q + 13) = 0.01q^3 - 0.6q^2 + 13q.\] This represents the total cost of producing \(q\) units.

Step 3: Deriving Marginal Cost

Marginal cost, \(MC(q)\), is the derivative of the total cost function \(C(q)\): \[MC(q) = \frac{d}{dq}(0.01q^3 - 0.6q^2 + 13q) = 0.03q^2 - 1.2q + 13.\] This function provides the cost of producing one additional unit at quantity \(q\).

Step 4: Finding Minimum Marginal Cost

To find the minimum marginal cost, set \(MC'(q) = 0\). Compute \(MC'(q) = 0.06q - 1.2\) and solve \(0.06q - 1.2 = 0\), giving \(q = 20\). Use \(MC''(q) = 0.06 > 0\) to confirm it's a minimum. Thus, minimum marginal cost occurs at \(q = 20\).

Step 5: Analyzing Average Cost for Minimum

The average cost is minimized where \(\frac{da}{dq} = 0\). Compute \(\frac{da}{dq} = 0.02q - 0.6\) and solve \(0.02q - 0.6 = 0\), giving \(q = 30\). Substitute in \(a(q)\) to find the lowest average cost \(a(30) = 4\).

Step 6: Evaluating Marginal Cost at q=30

Substitute \(q = 30\) into the marginal cost function: \[MC(30) = 0.03(30)^2 - 1.2(30) + 13 = 4.\] At \(q = 30\), the marginal cost equals the minimum average cost, confirming the average cost is minimized where \(MC(q) = a(q)\), meaning production is efficient.

Key Concepts

Average Cost Function

The average cost function, given by the equation \(a(q) = 0.01q^2 - 0.6q + 13\), is a tool that provides insight into how the cost per unit changes with varying levels of production, \(q\). It helps businesses understand how efficiently they are producing goods.

• The term \(0.01q^2\) represents the increase in cost as production levels change, which captures the complexity of scaling production.
• The term \(-0.6q\) portrays a reduction per unit cost as initial investments pay off with higher production.
• The constant \(13\) reflects fixed costs that do not change with output, like rent or salaries.

To find the level of production that minimizes average cost, we studied the derivative of this function, which signifies the point where producing one more unit would start increasing the average cost per item.

Total Cost Function

The total cost function, denoted as \(C(q)\), represents the overall cost of producing \(q\) goods. It is obtained by multiplying the average cost function by the quantity produced. In our scenario, the total cost function becomes:\[C(q) = q(0.01q^2 - 0.6q + 13) = 0.01q^3 - 0.6q^2 + 13q.\]
This function allows businesses to understand their whole cost structure, which includes both variable costs (like materials and production hours) and fixed costs (like utilities and leases).
By analyzing the total cost function, companies can make informed decisions about scaling their production or identifying the optimal level of output that improves profitability.

Derivative

The derivative is a mathematical tool that is crucial in economics for finding rates of change. In the context of cost functions, taking the derivative helps find the marginal cost, which shows us how the total cost changes as more units are produced.For the total cost function \(C(q) = 0.01q^3 - 0.6q^2 + 13q\), the derivative gives us the marginal cost function: \[MC(q) = \frac{d}{dq} (0.01q^3 - 0.6q^2 + 13q) = 0.03q^2 - 1.2q + 13.\]
This derivative tells us the cost of producing just one additional unit and can be used to optimize production levels to ensure maximum efficiency and profitability.

Minimum Marginal Cost

To locate the minimum marginal cost, we focus on where the derivative of the marginal cost function, or the second derivative of the total cost function, equals zero. Solving for \(q\) in \(MC'(q) = 0 \) gives us:

• Compute the derivative: \(MC'(q) = 0.06q - 1.2\).
• Set the derivative to zero: \(0.06q - 1.2 = 0\).
• Solve for \(q\): \(q = 20\) tells us the production level for minimum marginal cost.
• Confirm the minimum by checking \(MC''(q) = 0.06\), which is greater than zero, indicating a minimum point.

At \(q = 20\), we achieve the lowest additional cost for producing one more unit, meaning the production operates at an efficient stage where costs begin increasing again if surpassed.