Problem 68 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

MARGINAL COST At a certain factory, the marginal cost is \(6(q-5)^{2}\) dollars per unit when the level of production is \(q\) units. By how much will the total manufacturing cost increase if the level of production is raised from 10 to 13 units?

Step-by-Step Solution

Answer

The total manufacturing cost will increase by $774.

Step 1: Understand Marginal Cost

The marginal cost function given is: \(MC(q) = 6(q-5)^{2} \). This represents the cost of producing one more unit at a production level of \(q \) units.

Step 2: Set Up the Integral

To find the increase in total manufacturing cost when production increases from 10 to 13 units, integrate the marginal cost function from \(q = 10 \) to \(q = 13 \). We need to compute: \(\text{Increase in cost} = \int_{10}^{13} 6(q-5)^2 \, dq \)

Step 3: Compute the Integral

Evaluate the definite integral: \(\text{Increase in cost} = 6 \int_{10}^{13} (q-5)^2 \, dq \)

Step 4: Simplify the Integrand

Use substitution where \(u = q - 5 \), and thus \(du = dq \)

Step 5: Change the Limits of Integration

Rewrite the limits of integration in terms of \(u \). When \(q = 10 \, u = 5 \). When \(q = 13 \, u = 8 \).

Step 6: Evaluate the Definite Integral

Perform the integration: \(\text{Increase in cost} = 6 \int_{5}^{8} u^2 \, du \) which results in \(6 \left[ \frac{u^{3}}{3} \right]_{5}^{8} \)

Step 7: Calculate the Result

Compute the definite integral: \(6 \left(\frac{8^3}{3} - \frac{5^3}{3}\right) = 6 \left(\frac{512}{3} - \frac{125}{3}\right) = 6 \left(\frac{387}{3}\right) = 6 \left(129\right) = 774 \)

Key Concepts

Integral Calculus

Integral calculus is a fundamental branch of calculus focusing on the concept of integration. Integration helps determine areas under curves, volumes, central points, and much more. At its core, it involves summing up an infinite number of infinitesimal quantities. In most practical applications, you will come across two types of integrals: indefinite integrals and definite integrals. Indefinite integrals represent a family of functions, while definite integrals give a specific numerical value.

Definite Integral

A definite integral represents the net area under a curve within a specific interval. In mathematical notation, it's expressed as: \( \int_{a}^{b} f(x) \, dx \ \) This reads as 'the integral from \( a \) to \( b \) of \( f(x) \) with respect to \( x \).' Here, \( a \) and \( b \) are the limits of integration. Calculating a definite integral involves:

• Choosing the function to integrate.
• Setting the limits \( a \) and \( b \).
• Applying the fundamental theorem of calculus, which connects differentiation and integration.

To make this concrete, consider the integral in our exercise: \( \int_{10}^{13} 6(q-5)^2 \, dq \ \) Here, we integrated the marginal cost function within the given production levels to find out the increase in the total cost.

Production Cost Analysis

Production cost analysis is vital for businesses to understand their cost dynamics. One crucial concept is marginal cost, which denotes the cost of producing one additional unit. This helps in decision-making, especially in optimizing production levels. Using calculus, particularly integration, allows us to sum up the marginal costs to find the total production cost over an interval. For example, in our problem, we started with: \( MC(q) = 6(q-5)^2 \ \) Integrating this from \( q = 10 \) to \( q = 13 \) gave us the total increase in manufacturing cost due to increased production. This provides actionable insights into how much additional production will cost.