Problem 51 - University Calculus: Early Transcendentals

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 $$n=\frac{a}{x-c}+b(100-x)$$ 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 \(x = \frac{100b + c}{2b}\).

Step 1: Understand the Components of Profit

The profit from selling each backpack is the selling price, \(x\), minus the manufacturing cost, \(c\). Therefore, profit per backpack is \(x - c\).

Step 2: Write the Profit Function

To find the total profit, multiply the profit per backpack by the number sold, \(n\). Thus, the profit \(P(x)\) is given by \(P(x) = (x-c)\left(\frac{a}{x-c} + b(100-x)\right)\).

Step 3: Simplify the Profit Function

Expand the expression within the brackets and simplify: \((x-c)\left(\frac{a}{x-c} + b(100-x)\right) = a + b(x-c)(100-x)\). This will simplify further as \(P(x) = a + b(100x - x^2 - 100c + xc)\).

Step 4: Simplify Further the Profit Function

After distributing, combine like terms into a quadratic form: \(P(x) = a + 100bx - bx^2 - 100bc + bxc\). This becomes a quadratic function of the form \(-bx^2 + (100b+c)x + (a - 100bc)\).

Step 5: Find the Vertex of the Quadratic Function

Since the quadratic function is of the form \(-bx^2 + (100b+c)x + (a - 100bc)\), find the vertex using the formula \(x = -\frac{B}{2A}\) for the maximum value of a downward-facing parabola. Here, \(A = -b\) and \(B = 100b + c\).

Step 6: Calculate the Optimal Selling Price

Calculate the vertex to determine the selling price that maximizes profit: \(x = \frac{100b + c}{2b}\).

Key Concepts

Quadratic Functions

Quadratic functions form the basis of many mathematical problems, especially those dealing with optimization. A quadratic function is a polynomial with a degree of two, typically written in the form \( f(x) = ax^2 + bx + c \) where \( a \), \( b \), and \( c \) are constants.

The graph of a quadratic function is a parabola, which has a specific shape that curves either upwards or downwards.

This curvature is determined by the sign of the \( a \) coefficient:

• If \( a > 0 \), the parabola opens upwards.
• If \( a < 0 \), the parabola opens downwards.

Understanding the nature of quadratic functions helps in finding critical points like maxima and minima, which are vital for problems such as profit maximization.

Vertex of a Parabola

To locate a parabola's highest or lowest point, one must identify its vertex. The vertex of a parabolic function \( ax^2 + bx + c \) can be found using the formula \( x = -\frac{B}{2A} \).

This equation provides the \( x \)-coordinate of the vertex, and in the context of profit maximization, it indicates the selling price that yields maximum profit.

Once you have the \( x \), plug it back into the original quadratic equation to find the maximum profit value or the lowest cost.

Knowing the vertex is crucial because:

• It tells you where the maximum or minimum value of the function occurs.
• It simplifies the process of finding optimal strategies in economic problems.

Price Optimization

Price optimization is a strategic process that involves determining the best selling price to maximize revenue or profit. In mathematical terms, it utilizes functions to model the relationship between pricing and sales.

The main goal is to find the price that maximizes the total profit, which often relies on understanding the vertex of the quadratic profit function.

In this context, price optimization takes into account factors like:

• How changes in price affect the quantity sold.
• The cost of production impacting the profit per unit.
• Consumer demand variability.

Using quadratic functions, we can derive a formula that helps to pinpoint the ideal price point, ensuring that businesses do not overprice or underprice their products.

Profit Function

The profit function in business refers to the mathematical representation of how profits are determined based on costs and revenues. In this exercise, the profit function \( P(x) = (x-c)\left(\frac{a}{x-c} + b(100-x)\right) \) starts by considering:

• The number of units sold.
• The selling price per unit minus the production cost per unit.

This expression ultimately revolves around the concept that a higher selling price does not necessarily guarantee higher profits if the quantity sold decreases significantly.

Maximizing the profit function requires simplifying it into a quadratic form. Then, finding the vertex to ensure that the company sets the optimal price that maximizes profit while covering costs.

Calculus Optimization

Calculus optimization addresses finding maximum and minimum values of functions through derivatives. In the context of pricing and profit, calculus helps identify the optimal price point where a company's profit reaches its peak.

Using differentiation, we can:

• Take the first derivative of the profit function to find critical points, often where the slope equals zero.
• Use the second derivative test to ensure these points represent a maximum or a minimum.
• Apply this for complex profits in quadratic functions, easing identification of the best price for maximum profit.

In this exercise, after obtaining the quadratic profit function, identifying its vertex through calculus optimization further validates the precision of finding the most profitable price point.