Problem 31 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

A manufacturer can produce sunglasses at a cost of \(\$ 5\) apiece and estimates that if they are sold for \(x\) dollars apiece, consumers will buy \(100(20-x)\) sunglasses a day. At what price should the manufacturer sell the sunglasses to maximize profit?

Step-by-Step Solution

Answer

Sell the sunglasses at \(12.5\) dollars each to maximize profit.

Step 1: Define the Profit Function

Profit is defined as total revenue minus total cost. If the manufacturer sells the sunglasses for \( x \) dollars each, the number of sunglasses sold is \( 100(20 - x) \). The revenue, therefore, is \( x \times 100(20 - x) = 100x(20 - x) \). The cost of producing \( 100(20 - x) \) sunglasses is \( 5 \times 100(20 - x) = 500(20 - x) \). The profit function \( P(x) \) is then: \[ P(x) = 100x(20 - x) - 500(20 - x) \] = \[ P(x) = 2000x - 100x^2 - 10000 + 500x \] = \[ P(x) = -100x^2 + 2500x - 10000 \]

Step 2: Find the Critical Points

To find the maximum profit, differentiate the profit function \( P(x) \) with respect to \( x \) and set the derivative equal to zero: \[ P'(x) = -200x + 2500 \] Solving \( P'(x) = 0 \): \[ -200x + 2500 = 0 \] \[ 200x = 2500 \] \[ x = 12.5 \]

Step 3: Determine the Nature of the Critical Point

To confirm that \( x = 12.5 \) maximizes the profit, use the second derivative test. The second derivative of \( P(x) \) is: \[ P''(x) = -200 \] Since \( P''(x) \) is negative, the critical point at \( x = 12.5 \) corresponds to a maximum.

Step 4: State the Optimal Price

The manufacturer should sell the sunglasses at \( x = 12.5 \) dollars each to maximize profit.

Key Concepts

Profit Function

The profit function is essential when determining how a business can maximize its financial gain. In this exercise, the profit function is defined as total revenue minus total costs.

To break it down:

• Total revenue is the amount of money earned from selling the product.
• Total cost is the sum spent to produce the product.

In this specific example, the revenue depends on the selling price per unit and the number of units sold. The cost depends on the number of units produced and the cost per unit.

By subtracting the total cost from the total revenue, we get the profit function. For instance, the profit function for the sunglasses example is: \[ P(x) = -100x^2 + 2500x - 10000 \] where x is the selling price per unit.

Differentiation in Calculus

Differentiation is a fundamental concept in calculus that helps us understand how a function changes. It's all about finding the rate at which one quantity changes with respect to another.

For the profit function, we use differentiation to find the rate of change of profit concerning the selling price, x.

The derivative of a function gives us the slope of the function at any point. By setting this derivative to zero, we can find the critical points where the function could have a maximum or minimum.

In our exercise, the first derivative of the profit function is: \[ P'(x) = -200x + 2500 \] Setting this to zero allows us to solve for x and find the critical points for the maximum profit.

Second Derivative Test

The second derivative test determines whether a critical point found using the first derivative is a maximum, minimum, or a saddle point. This is crucial when trying to maximize profit.

Here's how it works:

• Calculate the second derivative of the profit function.
• Evaluate the second derivative at the critical points found earlier.

If the second derivative at a critical point is negative, the function has a local maximum at that point. If it is positive, the function has a local minimum.

In our scenario, the second derivative of the profit function is: \[ P''(x) = -200 \] Since this is negative (\

Critical Points

Critical points are where the first derivative of a function equals zero or is undefined. These points are potential locations for the maximum and minimum values of the function.

To find these points for the profit function, we set the first derivative equal to zero and solve for x.

In our problem, solving the equation \[ -200x + 2500 = 0 \] gives us: \[ x = 12.5 \] This selling price is a critical point where we might achieve maximum profit.

Revenue and Cost Functions

The revenue and cost functions are crucial for the profit function. They help us understand how different price points affect profit.

For our sunglasses example:

• Revenue Function: This represents the total money earned. For x dollars per unit sold, the revenue is: \[ 100x(20 - x) \]
• Cost Function: This represents the total cost of production. Producing 100(20 - x) sunglasses at $5 each, the cost is: \[ 500(20 - x) \]

By substituting these into the profit function, we get the formula that tells us how profit changes with the selling price.