Problem 11 - Applied Calculus

Question

Let \(C(q)\) represent the cost and \(R(q)\) represent the revenue, in dollars, of producing \(q\) items. 5 (a) If \(C(50)=4300\) and \(C^{\prime}(50)=24,\) estimate \(C(52)\) (b) If \(C^{\prime}(50)=24\) and \(R^{\prime}(50)=35,\) approximately how much profit is earned by the \(51^{\text {st }}\) item? (c) If \(C^{\prime}(100)=38\) and \(R^{\prime}(100)=35,\) should the company produce the \(101^{\text {st }}\) item? Why or why not?

Step-by-Step Solution

Answer

(a) \( C(52) \approx 4348 \). (b) Profit for 51st item is \$11. (c) Do not produce the 101st item due to loss.

Step 1: Estimating Cost for C(52) using Derivative

Given that \( C(50) = 4300 \) and \( C'(50) = 24 \), we can use the tangent line approximation for small changes around \( q = 50 \). The formula for linear approximation is:\[ C(q + \Delta q) \approx C(q) + C'(q) \cdot \Delta q \]Here, \( \Delta q = 52 - 50 = 2 \). Thus, \[ C(52) \approx C(50) + C'(50) \times 2 \]Substitute the given values: \[ C(52) \approx 4300 + 24 \times 2 \]Calculate it:\[ C(52) \approx 4300 + 48 = 4348 \]

Step 2: Calculating Profit for the 51st Item

Given that the derivative of cost, \( C'(50) = 24 \), represents the cost change for each additional item, and the derivative of revenue, \( R'(50) = 35 \), represents the revenue change for each additional item, the profit for the \( 51^{\text{st}} \) item can be calculated as:\[ \text{Profit for 51st item} = R'(50) - C'(50) \]Substitute the given values:\[ \text{Profit for 51st item} = 35 - 24 = 11 \]

Step 3: Evaluating Production of the 101st Item

To decide whether to produce the \( 101^{\text{st}} \) item, compare the cost and revenue changes:- \( C'(100) = 38 \)- \( R'(100) = 35 \)Since the marginal cost \( C'(100) \) is greater than the marginal revenue \( R'(100) \), producing the \( 101^{\text{st}} \) item would result in a loss.Therefore, it is not profitable to produce the \( 101^{\text{st}} \) item.

Key Concepts

Understanding Marginal Cost

Marginal cost is a vital concept in economics and business. It represents the additional cost incurred from producing one more unit of a good or service. When discussing marginal cost, we are focusing on how much it costs to increase production by one unit. This is calculated through the derivative of the cost function, represented as \(C'(q)\). If you have a cost function \(C(q)\), by finding its derivative, you can determine how the cost changes with each additional item. For example, with \(C'(50) = 24\), it means each extra item produced after 50 costs an additional 24 dollars.
Applying marginal cost helps businesses decide how much to produce. If the cost of producing one more unit is higher than the revenue gained from selling it, it may not be worthwhile to increase production. Understanding this balance is critical for profit maximization, allowing businesses to make informed decisions on production levels.

Exploring the Revenue Function

The revenue function, denoted as \(R(q)\), shows the total revenue a company earns from selling \(q\) units of a product. The derivative of the revenue function, \(R'(q)\), tells us how revenue changes with each additional unit sold, which is known as marginal revenue.
Marginal revenue is a crucial figure in determining profit margins. If you know the rate at which revenue changes, you can predict future earnings more accurately. For example, with \(R'(50) = 35\), this indicates each additional item sold after 50 brings in an extra 35 dollars.
This information, combined with the marginal cost, allows businesses to analyze profitability. If \(R'(q) > C'(q)\), the company earns more from selling an additional item than it costs to produce it, signifying a profit. However, if \(R'(q) < C'(q)\), producing more would lead to losses. Thus, understanding the revenue function's derivative is essential for evaluating potential profits.

Tangent Line Approximation in Cost Estimation

Tangent line approximation is a mathematical technique used to estimate function values near a point using the value of the function and its derivative at that point. It relies on the linear approximation formula, \[C(q + \Delta q) \approx C(q) + C'(q) \cdot \Delta q,\] where \(\Delta q\) represents a small change from the initial point.
This method is handy when dealing with cost functions and needing quick estimates. By using the derivative \(C'(q)\), which represents the rate of change, you can predict changes in cost with minor production adjustments.
For instance, if \(C(50) = 4300\) and \(C'(50) = 24\), to find \(C(52)\), you approximate it as \(C(50) + 24 \times 2\), resulting in 4348 dollars.
Using tangent line approximation simplifies complex calculations, providing a fast and efficient way to gauge impacts of small adjustments on production and costs.