Problem 65 - Calculus Early Transcendentals: Pearson New International Edition

Question

The XYZ Company manufactures wicker chairs. With its present machines, it has a maximum yearly output of 500 units. If it makes \(x\) chairs, it can set a price of \(p(x)=200-0.15 x\) dollars each and will have a total yearly cost of \(C(x)=5000+6 x-0.002 x^{2}\) dollars. The company has the opportunity to buy a new machine for \(\$ 4000\) with which the company can make up to an additional 250 chairs per year. The cost function for values of \(x\) between 500 and 750 is thus \(C(x)=9000+6 x-0.002 x^{2} .\) Basing your analysis on the profit for the next year, answer the following questions. (a) Should the company purchase the additional machine? (b) What should be the level of production?

Step-by-Step Solution

Answer

Purchase the new machine and produce 655 chairs for maximum profit.

Step 1: Understand the Profit Function

The profit function is given by profit = revenue - cost. Here, revenue is calculated as the price per unit multiplied by the number of units sold, i.e., \( R(x) = p(x) \times x = (200 - 0.15x) \times x \). Therefore, the profit function is \( P(x) = R(x) - C(x) \).

Step 2: Calculate Initial Profit Function Without New Machine

For production output \( x \) not exceeding 500, the profit function is calculated as follows:\[ P(x) = (200x - 0.15x^2) - (5000 + 6x - 0.002x^2) \]Simplify the equation:\[ P(x) = 194x - 0.148x^2 - 5000 \]

Step 3: Calculate Profit Function With New Machine

For production \( x \) up to 750, the profit function is:\[ P(x) = (200x - 0.15x^2) - (9000 + 6x - 0.002x^2) \]Simplify this:\[ P(x) = 194x - 0.148x^2 - 9000 \]

Step 4: Find Maximum Profit Without New Machine

To find the maximum profit, take the derivative of \( P(x) = 194x - 0.148x^2 - 5000 \) and set it to zero:\[ \frac{dP}{dx} = 194 - 0.296x = 0 \]Solving gives \( x \approx 655.41 \), but limited to 500, the point of analysis to consider is \( x = 500 \).Calculate \( P(500) \):\[ P(500) = 194 \times 500 - 0.148 \times 500^2 - 5000 \approx \$42400 \]

Step 5: Find Maximum Profit With New Machine

Take the derivative of \( P(x) = 194x - 0.148x^2 - 9000 \) and set to zero:\[ \frac{dP}{dx} = 194 - 0.296x = 0 \]Solving gives \( x \approx 655.41 \), use this point:Calculate \( P(655) \):\[ P(655) = 194 \times 655 - 0.148 \times 655^2 - 9000 \approx \$42691.50 \]

Step 6: Conclusion and Recommendation

Comparing profits, without new machine, the profit for producing 500 chairs is approximately \\(42400. With the new machine, maximum profit at producing 655 chairs is about \\)42691.50. Thus, purchasing the machine offers slightly more profit, making it worthwhile to purchase and produce 655 chairs for maximum profit.

Key Concepts

Cost Function

The cost function is fundamental in understanding how much it costs to produce goods in manufacturing. For XYZ Company, making wicker chairs involves fixed and variable costs. The cost function describes these costs mathematically and is denoted as follows:

• For production levels up to 500 chairs, the cost function is: \(C(x) = 5000 + 6x - 0.002x^2\).
• For production between 500 and 750 chairs, it changes to: \(C(x) = 9000 + 6x - 0.002x^2\).

Fixed costs are costs that do not change with the level of output (e.g., \\(5000 or \\)9000), while variable costs change with production levels. The linear term \(6x\) represents variable costs proportional to the number of chairs made. The quadratic term \(-0.002x^2\) accounts for more complex dependencies, usually inefficiencies or savings in scale. Understanding the cost function lets a company optimize production while considering costs.

Revenue Function

Revenue functions calculate the total income from product sales. For XYZ Company, the price per chair depends on the quantity produced, affecting potential revenue. The revenue function is expressed as: \(R(x) = (200 - 0.15x) imes x\). Here:

• \(200\) is the initial price per chair.
• \(-0.15x\) captures how price decreases as more chairs are produced, a common situation when increasing output requires reducing prices to increase sales.

This model assumes the law of demand, where higher supplies can lower prices or require discounts. Hence, understanding the revenue function assists in deciding optimal production levels that maximize earnings.

Derivative for Optimization

In mathematics, derivatives help locate maximum or minimum points on curves, crucial for optimization. For XYZ Company, optimizing profit involves using derivatives of the profit function, \(P(x) = R(x) - C(x)\). The derivative of the profit function, \(\frac{dP}{dx}\), is set to zero to find the maximum profit point. For each scenario (with and without the new machine), calculate the derivative:

• Without Machine: \(\frac{dP}{dx} = 194 - 0.296x\)
• With Machine: \(\frac{dP}{dx} = 194 - 0.296x\)

Solving \(\frac{dP}{dx} = 0\) reveals the chair production levels that maximize profit; around 655 chairs with the new machine. Derivatives are the mathematical tool that bridge cost and revenue functions for effective profit maximization.

Marginal Analysis

Marginal analysis examines the benefits of production changes one unit at a time. In profit maximization, understanding how small changes in production affect profit is essential. For XYZ Company:

• Marginal Cost is the cost of producing one more chair, derived from the cost function's derivative.
• Marginal Revenue is the income from selling one more chair, derived from the revenue function's derivative.

When marginal revenue equals marginal cost, profits are optimized. This principle dictates that producing up to the point where additional units cease to add to profits is most efficient. Hence, the recommendation to produce 655 chairs stems from this marginal analysis approach, where additional chairs still contribute to profit until this equilibrium point.