Problem 62 - Calculus an Applied Approach

Question

Managing a Store You are managing a store and have been adjusting the price of an item. You have found that you make a profit of \(\$ 50\) when 10 units are sold, \(\$ 60\) when 12 units are sold, and \(\$ 65\) when 14 units are sold. (a) Fit these data to the model \(P=a x^{2}+b x+c .\) (b) Use a graphing utility to graph \(P .\) (c) Find the point on the graph at which the marginal profit is zero. Interpret this point in the context of the problem.

Step-by-Step Solution

Answer

After solving the system of equations and fitting the data into the model, the quadratic function should be obtained. The graph of the function will be a parabola. The point at which marginal profit is zero will be the maximum point of the graph, representing the number of units that should be sold for maximum profit. By interpreting this point, we find that increasing or decreasing the number of units sold from this point will not increase the profit, but decrease it instead.

Step 1: Fitting the data into the model

Set up a system of equations to find \(a\), \(b\), and \(c\). Substitute \(x = 10, 12, 14\) into the model \(P = ax^{2} + bx + c\) to get three equations. The profit for each is given as \(50, 60, 65\), respectively. Solve the resulting system of equations.

Step 2: Graphing the quadratic function

After obtaining the values of \(a\), \(b\), and \(c\), replace them in the quadratic function \(P = ax^{2} + bx + c\). Then plot the quadratic function using graphing utility keeping \(x\) on the horizontal axis (representing units sold) and \(P\) on the vertical axis (representing profit in dollars). You should get a parabola shape.

Step 3: Find the point of marginal profit being zero

To find the marginal profit, take the derivative of the profit function. Because our function is quadratic, the derivative will be a linear function. Set the derivative equal to zero and solve for \(x\), which points to the number of units sold at which the marginal profit becomes zero. The \(y\) coordinate is the profit at this point. This point corresponds to the maximum (peak of the parabola) of the profit function.

Step 4: Interpretation

The x-coordinate of the point at which marginal profit is zero corresponds to the number of units that yield maximum profit. Increasing or decreasing the number of units sold from this point will not increase the profit, but decrease it instead.

Key Concepts

Marginal Profit

Marginal profit is a critical concept in business and economics, representing the additional profit gained by selling one more unit of a product. To find the marginal profit in a mathematical model, we differentiate the profit function, which in our case is a quadratic function. Once the profit function, referred to as \( P \), is differentiated with respect to the number of units \( x \), we obtain a linear function that indicates the rate of change of profit. If this derivative is set to zero, the result gives us the number of units sold for which any additional units will not increase profit. In fact, beyond this point, selling more units may actually decrease the total profit, signifying an optimized level of production and sales.

For the store manager in the exercise, calculating the marginal profit and finding the point where it becomes zero is essential for maximizing profit. It enables the manager to determine the most profitable number of units to sell before the cost of producing the next unit exceeds the revenue it would generate. Understanding this concept is crucial for making informed pricing and production decisions.

Quadratic Functions

Quadratic functions are an integral part of algebra, characterized by an equation of the form \( P(x) = ax^2 + bx + c \), where \( a, b, \) and \( c \) are constants with \( a \) not equal to zero. These functions graph to a curve called a parabola, which has a distinct 'U' shape. The shape can open upwards or downwards depending on the sign of the coefficient \( a \). When modeling profits, quadratic functions can describe a wide range of scenarios, from product pricing to revenue forecasts.

In our store management example, the quadratic function models the relationship between units sold and profits. It's important to note that while profits may initially increase with more sales, due to increasing costs or market saturation, there could be a point at which additional sales lead to diminishing returns or even losses. By studying quadratic functions in real-world contexts, students can gain practical insights into business operations.

Solving Systems of Equations

In many mathematical and real-world scenarios, we encounter systems of equations—sets of two or more equations that are solved together because they share variables. Understanding how to solve these systems is vital in multiple fields, including business, engineering, and science. There are various methods to solve such systems, including substitution, elimination, and matrix operations.

In the textbook example, we see a system of equations established by substituting known values of units sold and profits into the quadratic model. Solving this system allows one to find the specific model that fits the data points provided by the store's profit experience. Mastery of solving systems of equations helps students not just in academic pursuits, but also equips them with the analytical skills necessary for solving complex problems where multiple variables and constraints are in play.

Graphing Quadratic Equations

Graphing is a powerful tool for visualizing mathematical concepts, and graphing quadratic equations provides insights into the nature of the solutions represented by these equations. A graphed quadratic equation takes the shape of a parabola, which can aid in understanding the maximum or minimum points, the direction of opening, and the axis of symmetry.

Using a graphing utility, as suggested in Step 2 of the solution, helps students to observe the parabolic trend of profit as units sold increase. This visual aid can also demonstrate the practical effect of pricing strategies and can be used to predict future profits or losses. The graphical approach to understanding quadratic equations is especially beneficial for visual learners, allowing them to 'see' the abstract equations taking shape in a concrete form. It reinforces analytical skills through visual interpretation, which is key to many aspects of problem-solving in various fields.