Problem 55 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

The selling price \(x\) that maximizes profit is found by solving \(P'(x) = 0\).

Step 1: Define the Profit Function

Profit is the total revenue minus the total cost. We begin by finding expressions for both revenue and cost. The total revenue, when selling \(n\) backpacks at \(x\) dollars each, is \(R(x) = xn\). The total cost, for manufacturing \(n\) backpacks at a cost of \(c\) dollars each, is \(C(x) = cn\). So, the profit function \(P(x)\) becomes:\[P(x) = R(x) - C(x) = xn - cn.\]

Step 2: Substitute for n

Using the given formula for \(n\), substitute it into our profit equation:\[P(x) = x\left( \frac{a}{x-c} + b(100 - x) \right) - c\left( \frac{a}{x-c} + b(100 - x) \right).\]Simplifying, this results in:\[P(x) = \frac{ax}{x-c} + 100b(x - c) - bx^2 - \frac{ac}{x-c}.\]

Step 3: Differentiate the Profit Function

To find the maximum profit, differentiate \(P(x)\) with respect to \(x\). This is done using the quotient rule and chain rule. First find:\[P'(x) = \frac{d}{dx}\left( \frac{ax}{x-c} \right) + \frac{d}{dx}(100bx) - \frac{d}{dx}(bx^2) - \frac{d}{dx}\left( \frac{ac}{x-c} \right).\]

Step 4: Set the Derivative Equal to Zero

For maximum profit, set \(P'(x) = 0\) and solve for \(x\). This is done by solving:\[0 = \frac{d}{dx}\left( \frac{ax}{x-c} \right) + 100b - 2bx - \frac{d}{dx}\left( \frac{ac}{x-c} \right).\]Solve this resulting equation for \(x\) to find the critical points.

Step 5: Verify the Maximum

Use the second derivative test, \(P''(x)\), to confirm if the critical point is a maximum. Compute the second derivative \(P''(x)\) and evaluate it at the critical value. If \(P''(x) < 0\), the point is a maximum.

Key Concepts

Profit Function

Understanding the concept of profit is key to maximizing it. Profit is essentially the money you make after deducting all the costs associated with producing and selling a product. In our problem, backpacks are the product.
To create a profit function, we start with revenue and cost:

• Revenue: This is the income from selling the backpacks, calculated as the selling price per backpack multiplied by the number of backpacks sold, i.e., \( R(x) = xn \).
• Cost: This includes all money spent on producing the backpacks, formulated as the manufacturing cost per backpack multiplied by the number of backpacks, i.e., \( C(x) = cn \).

By subtracting total cost from total revenue, we get the profit function: \( P(x) = R(x) - C(x) = xn - cn \). This function allows us to understand how profit changes with different selling prices \( x \).

Derivative

To maximize profit, we need to investigate how changes in selling price \( x \) affect the profit function. Here’s where derivatives come into play. A derivative essentially tells us the rate at which one quantity changes relative to another.
For the profit function \( P(x) \), its derivative \( P'(x) \) shows how quickly profit changes as selling price \( x \) shifts. To find \( P'(x) \), we apply calculus tools like the quotient rule and the chain rule. The quotient rule is particularly useful when dealing with fractions, while the chain rule helps with differentiating composite functions.
Using these rules, we derive:

• \( P'(x) = \frac{d}{dx}\left( \frac{ax}{x-c} \right) + 100b - 2bx - \frac{d}{dx}\left( \frac{ac}{x-c} \right) \)

By calculating this derivative, we can determine where the profit changes direction, which can indicate a maximum profit point.

Critical Points

Critical points play a crucial role in determining where maximum profit occurs. A critical point is found where the derivative of a function is zero or undefined. For the profit maximization, we solve the equation \( P'(x) = 0 \).
This step involves setting our derived formula to zero and solving for \( x \):

• \( 0 = \frac{d}{dx}\left( \frac{ax}{x-c} \right) + 100b - 2bx - \frac{d}{dx}\left( \frac{ac}{x-c} \right) \)

This calculation helps us pinpoint the specific price at which profit doesn’t increase or decrease — a potential sweet spot for maximizing profit.
However, identifying the critical point alone isn’t enough; we must confirm whether it gives us a maximum or minimum profit. This is where the Second Derivative Test comes in handy.

Second Derivative Test

The Second Derivative Test is a method for confirming the nature of a critical point. After solving \( P'(x) = 0 \) and finding the critical points, we compute the second derivative \( P''(x) \) of the profit function.
Here's how it works:

• If \( P''(x) > 0 \), the critical point is a local minimum.
• If \( P''(x) < 0 \), the critical point is a local maximum.
• If \( P''(x) = 0 \), the test is inconclusive, and further analysis is needed.

For our exercise, once we determine \( P''(x) \) at the critical value and find it to be less than zero, we conclude that the selling price at this point indeed maximizes the profit. This guarantees that we're suggesting a price that brings the maximum financial return.