Problem 43 - Thomas Calculus

Question

It costs you \(c\) dollars each to manufacture and distribute back- packs. If the backpacks sell at \(x\) dollars each, the number sold is given by $$ n=\frac{a}{x-c}+b(100-x) $$ where \(a\) and \(b\) are positive constants. What selling price will bring a maximum profit?

Step-by-Step Solution

Answer

The selling price that maximizes profit is found by solving the derivative of the profit function, \(P(x)\), and verifying the maximum with the second derivative test.

Step 1: Define Profit Function

The profit function is determined by subtracting the total cost from the total revenue. In this case, total revenue is the selling price times the number sold, and total costs are the number sold times the cost to manufacture. Thus, the profit function is given by:\[ P(x) = x \left( \frac{a}{x-c} + b(100-x) \right) - c \left( \frac{a}{x-c} + b(100-x) \right) \]

Step 2: Simplify Profit Function

Simplify the profit function by distributing the terms and combining like terms:\[ P(x) = x \cdot \frac{a}{x-c} + b(100-x)x - c \cdot \frac{a}{x-c} - cb(100-x) \]

Step 3: Differentiate Profit Function

Calculate the first derivative of the profit function \(P(x)\) with respect to \(x\) to find its critical points. This derivative, \(P'(x)\), will help identify the selling price that maximizes profit.\[ P'(x) = \text{Derivative of } x \left( \frac{a}{x-c} + b(100-x) \right) - c \left( \frac{a}{x-c} + b(100-x) \right) \]

Step 4: Solve for Critical Points

Set the first derivative, \(P'(x)\), equal to zero to find the critical points:\[ P'(x) = 0 \]

Step 5: Determine Maximum Profit

Use the second derivative test or check the sign changes in the first derivative to confirm whether the critical points found deliver maximum profit. The value of \(x\) at which \(P(x)\) is a maximum is the optimal selling price.

Key Concepts

Derivative

A derivative is a fundamental tool in calculus. It measures how a function's output changes as its input changes. Think of it as a function's "instantaneous rate of change," or simply how quickly or slowly the function is increasing or decreasing at any point.

When dealing with profit maximization, derivatives play a crucial role. By differentiating a profit function with respect to the selling price, we gain insights into how small changes in price affect profit. This provides valuable data for identifying optimal pricing strategies.

The process involves taking the first derivative of the profit function. By understanding the derivative, we can find **critical points**, which indicate potential maximum or minimum profit levels.

First Derivative Test

The first derivative test is a method for determining the local maximum or minimum points of a function. Once you've calculated the first derivative of a profit function, set it to zero and solve for the variable to find critical points.

Here's what you need to do:

• Calculate the first derivative: This requires basic differentiation skills, often involving the product and quotient rules.
• Set the derivative equal to zero: Solve for the price variable to find critical values.
• Evaluate the sign changes: Pick values from intervals around the critical points to see if the derivative changes from positive to negative (indicating a maximum).

The selling price that results in the positive to negative sign change in the derivative identifies the price that potentially maximizes profit.

Profit Function

The profit function is at the heart of any business's strategy. It is formulated by subtracting total costs from total revenues, representing the actual profit made.

For instance, if you have a selling price of \(x\) per backpack, the revenue is the number of backpacks sold multiplied by \(x\). Total cost would be the cost per backpack times the quantity sold.

In our example, the profit function is \[ P(x) = x \frac{a}{x-c} + b(100-x)x - c \frac{a}{x-c} - cb(100-x) \]This equation shows a delicate balance, where changes in selling price affect both the number sold and, consequently, the profit.

Optimization techniques, using derivatives, highlight the selling price delivering the highest possible profit while considering production and overhead costs.

Second Derivative Test

The second derivative test confirms whether a critical point from the first derivative test is a maximum or a minimum.

Follow these steps:

• Compute the second derivative of the profit function.
• Plug the critical points from the first derivative into this second derivative.
• Assess the results: If the second derivative is negative at a critical point, it indicates a local maximum, confirming that it's the optimal price for maximum profit.

This test provides assurance that the selling price you choose indeed optimizes the profit.

Using both the first and second derivative tests gives a robust method for profit maximization, ensuring not just a mathematically sound solution but a practically beneficial one.