Problem 32 - A Graphical Approach to College Algebra

Question

Solve each problem. Maximizing Revenue Suppose the revenue \(R\) in thousands of dollars that a company receives from producing \(x\) thousand MP3 players is \(R(x)=x(40-2 x)\) (a) Evaluate \(R(2)\) and interpret the result. (b) How many MP3 players should the company produce to maximize its revenue? (c) What is the maximum revenue?

Step-by-Step Solution

Answer

(a) $72,000$ when producing 2,000 players. (b) 10,000 players for maximum revenue. (c) Maximum revenue is $200,000$.

Step 1: Evaluating Revenue at 2

To find the revenue when the company produces 2 thousand MP3 players, substitute \(x = 2\) into the revenue function: \[ R(2) = 2(40 - 2 \times 2) \]Simplify the expression:\[ R(2) = 2(40 - 4) = 2 \times 36 = 72 \]Thus, the revenue is 72 thousand dollars when producing 2 thousand MP3 players.

Step 2: Finding the Vertex of the Parabola

The revenue function \(R(x) = x(40 - 2x)\) can be rewritten as a quadratic equation: \[ R(x) = 40x - 2x^2 \]This is a quadratic function in the standard form \(ax^2 + bx + c\), where \(a = -2\) and \(b = 40\). To find the maximum revenue, we need to find the vertex. The vertex formula for the \(x\)-value is given by:\[ x = -\frac{b}{2a} \]Substituting the values of \(a\) and \(b\):\[ x = -\frac{40}{2(-2)} = \frac{40}{4} = 10 \]Therefore, the company should produce 10 thousand MP3 players to maximize revenue.

Step 3: Calculating Maximum Revenue

Now that we know the company should produce 10 thousand MP3 players, substitute \(x = 10\) into the revenue function to find the maximum revenue:\[ R(10) = 10(40 - 2 \times 10) \]Simplify the expression:\[ R(10) = 10(40 - 20) = 10 \times 20 = 200 \]Thus, the maximum revenue is 200 thousand dollars.

Key Concepts

Quadratic Functions

Quadratic functions are mathematical expressions that describe a particular kind of curve called a parabola. These functions take the form of \[ f(x) = ax^2 + bx + c \] where \( a \), \( b \), and \( c \) are constants, and \( x \) represents the variable. Quadratic functions are fundamental in algebra because they help us model certain types of relationships, such as the revenue of a company based on production levels.
To understand a quadratic function, look at its components:

• The \( a \) coefficient influences the direction of the parabola (upward if \( a > 0 \), downward if \( a < 0 \)).
• The \( b \) coefficient affects the tilt of the parabola.
• The \( c \) coefficient represents the y-intercept, or where the function crosses the y-axis.

The quadratic function in our problem, \[ R(x) = -2x^2 + 40x \], models how revenue changes as the number of items produced changes.
This function is essential for identifying the optimal production level for the highest revenue.

Vertex of a Parabola

Finding the vertex of a parabola is crucial, especially in revenue maximization problems. The vertex represents either the highest or lowest point of the parabola, depending on its direction.
In our problem, we deal with a downward-facing parabola since \( a = -2 \) is negative, indicating the vertex is the maximum point. The vertex formula \[ x = -\frac{b}{2a} \] allows us to find the \( x \)-value at the vertex. This value tells us the number of units that needs to be produced to achieve maximum revenue.
We substitute \( b = 40 \) and \( a = -2 \) into the formula:\[ x = -\frac{40}{2(-2)} = \frac{40}{4} = 10 \]Thus, the vertex occurs at \( x = 10 \). This means producing 10 thousand units brings the highest revenue. Understanding the vertex concept makes it easy to determine optimal solutions in business scenarios like ours.

Algebraic Expressions

Algebraic expressions are combinations of variables, numbers, and operators (like addition and multiplication). They form the building blocks of algebra, allowing us to represent mathematical models and relationships concisely.
The revenue function \[ R(x) = x(40 - 2x) \] is an algebraic expression. Let's break it down:

• The expression inside the parentheses \( (40 - 2x) \) shows how revenue per unit decreases as production \( x \) increases.
• It's multiplied by \( x \), representing quantity sold, to give the total revenue.

Using algebraic expressions makes solving complex problems possible. Here, we maximize revenue by simplifying the expression and applying known formulas.
This demonstrates why understanding algebra is vital: it provides tools to manipulate expressions and solve for unknowns efficiently.