Problem 69 - Precalculus with Limits

Question

REVENUE The revenue \(R\) (in dollars) generated by the sale of \(x\) units of a patio furniture set is given by \((x-106)^2 = -\dfrac{4}{5}(R-14,045)\). Use a graphing utility to graph the function and approximate the number of sales that will maximize revenue.

Step-by-Step Solution

Answer

Based on the graph, the x-coordinate of the vertex gives the number of units to be sold to maximize revenue, which can be approximated using the graphing tool.

Step 1: Rearrange the equation

The first step is to rearrange the given equation \((x-106)^2=-\dfrac{4}{5}(R-14,045)\) into a more familiar quadratic form. We can get rid of the negative sign by swapping the sides: \(-\dfrac{4}{5}(R-14,045)=(x-106)^2\). Multiply through by -5/4 to get rid of the fraction gives \(R=14,045-\dfrac{5}{4}(x-106)^2\). This now represents \(R\) as a function of \(x\), say \(R=f(x)\).

Step 2: Plot the graph

We can now plot the rearranged equation \(R=14,045-\dfrac{5}{4}(x-106)^2\). This is a quadratic function (quadratic equation), where the coefficient of \(x^2\) is negative so we are dealing with a parabola that opens downwards, meaning our solution would indeed be a maximum value. Using graphing software or a graphing calculator, plot the graph of the function \(R=14,045-\dfrac{5}{4}(x-106)^2\).

Step 3: Identify the vertex of the parabola

After plotting the graph, we can see that the graph takes the shape of a parabola as expected. The peak of the parabola which represents the maximum revenue (R) can be determined using the graphing utility or software. The x-coordinate of the vertex of the parabola is the number of units that ensures maximum revenue.

Key Concepts

Parabolic Function

A parabolic function is a special type of mathematical expression that creates a shape known as a parabola when graphed. This type of function comes from quadratic equations, which we will explore shortly. Parabolas can open upwards or downwards, depending on their coefficients. An upward-opening parabola looks like a "U," while a downward-opening parabola resembles an upside-down "U."

• Parabolas are symmetric along a vertical line known as the axis of symmetry.
• The highest or lowest point on a parabola is known as the vertex.
• They are used in a variety of applications, from physics to economics, as they can model the behavior of objects in motion or the relationship between variables, like in our revenue maximization problem.

Understanding parabolic functions and their graphs is essential for identifying key features such as the vertex and determining maximum or minimum values.

Quadratic Equation

A quadratic equation is a polynomial equation of the second degree. It typically has the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). The equation given in our exercise is transformed into a quadratic form to find the revenue function:

• The revenue equation is reshaped into \( R = 14,045 - \frac{5}{4}(x-106)^2 \), a clear quadratic form.
• Quadratic equations often result in a parabola when graphed, and here we see that since the \( (x-106)^2 \) term has a negative coefficient, the parabola opens downwards.
• This is crucial since the point of revenue maximization corresponds to the apex, or highest point, of the parabola.

By studying these equations, students can develop a deeper understanding of how different components affect the shape and position of the graph, since each term in the equation contributes to the overall behavior of the parabola.

Vertex of a Parabola

The vertex of a parabola is either its highest or lowest point, depending on its orientation. In our revenue problem, finding the vertex is essential because it corresponds to the maximum revenue generated, due to the downward-opening parabola. There are several key points to remember about the vertex:

• The formula for finding the vertex of a parabola \( y = ax^2 + bx + c \) is \( x = -\frac{b}{2a} \) to find the x-coordinate. However, for the form we have, \( R = 14,045 - \frac{5}{4}(x-106)^2 \), it is easier to inspect the expression.
• In this setup, the vertex occurs directly at \( x = 106 \), evidenced by the \( (x-106)^2 \) term.
• This x-coordinate gives us the number of units needed to maximize revenue, and the corresponding \( R \) value gives the maximum revenue itself.

Recognizing and calculating the vertex allows one to not only solve practical problems like maximizing revenue but also gain a comprehensive understanding of quadratic functions and their applications in real-world scenarios.