Problem 36 - Calculus an Applied Approach

Question

Marginal Profit The profit \(P(\text { in dollars) from selling } x\) units of a product is given by $$P=36,000+2048 \sqrt{x}-\frac{1}{8 x^{2}}, \quad 150 \leq x \leq 275$$ Find the marginal profit for each of the following sales. $$ \begin{array}{ll}{\text { (a) } x=150} & {\text { (b) } x=175 \quad \text { (c) } x=200} \\ {\text { (d) } x=225} & {\text { (e) } x=250 \quad \text { (f) } x=275}\end{array} $$

Step-by-Step Solution

Answer

The marginal profits for \(x = 150, 175, 200, 225, 250, 275\) are approximately 83.54, 77.32, 72.32, 68.27, 64.82, and 61.79 respectively.

Step 1: Find the derivative of the profit function

The derivative of the profit function gives the marginal profit. The derivatives of the functions \(\sqrt{x}\) and \(\frac{1}{x^2}\) are \(\frac{1}{2\sqrt{x}}\) and \(-\frac{2}{x^3}\) respectively. So, \(P'=2048 \cdot \frac{1}{2\sqrt{x}} + \frac{1}{8} \cdot 2x^{-3} = 1024\sqrt{x}^{-1} - \frac{1}{4}x^{-3} = \frac{1024}{\sqrt{x}} - \frac{1}{4x^3}\).

Step 2: Calculate the marginal profit for \(x = 150\)

Substitute 150 into the derivative function to get \(P'(150) = \frac{1024}{\sqrt{150}} - \frac{1}{4*150^3} = 83.54\).

Step 3: Calculate the marginal profit for \(x = 175\)

Substitute 175 into the derivative function to get \(P'(175) = \frac{1024}{\sqrt{175}} - \frac{1}{4*175^3} = 77.32\).

Step 4: Calculate the marginal profit for \(x = 200\)

Substitute 200 into the derivative function to get \(P'(200) = \frac{1024}{\sqrt{200}} - \frac{1}{4*200^3} = 72.32\).

Step 5: Calculate the marginal profit for \(x = 225\)

Substitute 225 into the derivative function to get \(P'(225) = \frac{1024}{\sqrt{225}} - \frac{1}{4*225^3} = 68.27\).

Step 6: Calculate the marginal profit for \(x = 250\)

Substitute 250 into the derivative function to get \(P'(250) = \frac{1024}{\sqrt{250}} - \frac{1}{4*250^3} = 64.82\).

Step 7: Calculate the marginal profit for \(x = 275\)

Substitute 275 into the derivative function to get \(P'(275) = \frac{1024}{\sqrt{275}} - \frac{1}{4*275^3} = 61.79\).

Key Concepts

Marginal Profit

Marginal profit is a crucial concept in calculus related to economics. It represents the additional profit a company earns from selling one more unit of a product. This helps companies decide optimal production levels. Also, it helps determine how changes in production affect overall profitability. Marginal profit can indicate whether increasing production will be beneficial or detrimental.

In a mathematical context, marginal profit is essentially the derivative of the profit function. By finding the derivative, we can determine how profit changes as production increases. This requires knowledge of differentiation, which is a foundational aspect of calculus.

Derivative

The derivative is a core concept in calculus. It measures how a function changes as its input changes. In simple terms, the derivative of a function provides the slope of the tangent line at any point on its curve.

In the context of marginal profit, the derivative of the profit function gives the rate at which profit is changing as more products are sold. This is crucial for understanding and forecasting business performance.

• The basic rules for finding derivatives include the power rule, product rule, quotient rule, and chain rule.
• For example, the derivative of \(\sqrt{x}\) is \(\frac{1}{2\sqrt{x}}\) and \(\frac{1}{x^2}\) is \(-\frac{2}{x^3}\).

The derivative provides valuable insights into the behavior of complex functions by revealing critical points and trends.

Profit Function

A profit function is a mathematical representation of a business's profitability. It depicts the relationship between the number of units sold and the profit generated. This function is essential for maximizing a company's financial results.

The profit function can be impacted by fixed and variable costs and revenue. It considers these elements to provide a comprehensive view of profitability at various output levels.

• In our exercise, the profit function is given by \(P = 36,000 + 2048 \sqrt{x} - \frac{1}{8x^2}\).
• The goal is to find how profit changes with different values of \(x\), the units sold.

Analyzing the profit function using calculus techniques like differentiation allows businesses to make informed financial decisions.

Applied Mathematics

Applied mathematics refers to the use of mathematical methods by different fields such as science, engineering, business, and industry. It focuses on practical problems, translating mathematical theories into real-world applications.

In our case, applied mathematics is used to understand the financial implications of production changes. Calculus, specifically derivatives, plays a significant role here.

• By applying differentiation techniques, we analyzed how marginal profit changes with various production levels.
• Understanding these changes supports strategic pricing and production planning.

This practical application of mathematics helps businesses optimize their operations and improve their financial outcomes.