Problem 74 - College Algebra and Calculus: An Applied Approach

Question

The demand equation for a product is \(p=60-0.0004 x\) where \(p\) is the price per unit and \(x\) is the number of units sold. The total revenue \(R\) for selling \(x\) units is given by \(R=x p\) How many units must be sold to produce a revenue of \(\$ 220,000 ?\)

Step-by-Step Solution

Answer

To produce a revenue of $220,000, 50,000 units must be sold.

Step 1: Insert the demand equation into the revenue equation

The demand price equation is \(p = 60 -0.0004x\), and the revenue equation is \(R = xp\). By replacing \(p\) in the revenue equation with our demand equation, we get a new revenue equation: \(R = x(60 - 0.0004x)\)

Step 2: Distribute and simplify this equation

By distributing, we get \(R = 60x - 0.0004x^2\)

Step 3: Equate this equation to the given revenue and solve for x

We know our desired revenue is \$220,000. After plugging this into our simplified equation and setting it equal to \$220,000, we get: \(220000 = 60x - 0.0004x^2\). This equation rearranges to become \(0.0004x^2 - 60x + 220000 = 0\). Solving this quadratic equation for \(x\) can be done using the quadratic formula \(x = [-b ± sqrt(b^2 - 4ac)] / 2a\)

Step 4: Solve the equation

Now plug the coefficients \(a = 0.0004\), \(b = -60\) and \(c = 220000\) into the quadratic formula. As price cannot be negative, so we choose the positive root. We get \(x = [60 + sqrt((60)^2 - 4*0.0004*220000)] / (2 * 0.0004)\). On solving, we find that \(x = 50000\)

Key Concepts

Understanding Quadratic Equations

A quadratic equation is a type of polynomial equation of degree 2. In standard form, it looks like this: \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(x\) is the variable we solve for. Quadratics are essential because they appear in various areas like physics, finance, and geometry.
In this exercise, we encounter a quadratic equation when calculating revenue. The equation comes from substituting the demand equation into the revenue formula. It takes the form \(0.0004x^2 - 60x + 220000 = 0\). Here:

• \(a = 0.0004\), the coefficient of \(x^2\)
• \(b = -60\), the coefficient of \(x\)
• \(c = 220000\), the constant term representing the desired revenue

To solve such an equation, the quadratic formula is commonly used: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). This formula helps us find the possible values of \(x\) that satisfy the equation.

Revenue Calculation and Optimization

Revenue calculation is crucial in determining how much income a business can generate by selling a certain number of products. Revenue is calculated by multiplying the number of units sold \(x\) by the price per unit \(p\), as given by \(R = xp\).
In the context of this problem, the revenue equation is adjusted by substituting the demand equation \(p = 60 - 0.0004x\) into it, resulting in a quadratic revenue equation: \(R = x(60 - 0.0004x)\). After simplification, it becomes \(R = 60x - 0.0004x^2\), showing a relationship between quantity sold and revenue.
Optimizing revenue involves finding the value of \(x\) that results in a specific revenue target, in this case, $220,000. Solving for \(x\) gives businesses insights into production levels required to meet revenue goals.

Algebraic Manipulation Techniques

Algebraic manipulation involves applying various mathematical techniques to simplify or solve equations. It requires skills in rearranging and simplifying terms for clarity and ease of solving.
In this problem, algebraic manipulation is demonstrated by plugging the demand equation \(p = 60 - 0.0004x\) into the revenue formula \(R = xp\) to form a more complete equation \(R = x(60 - 0.0004x)\).
Next, distribute \(x\) in the equation to expand it to \(R = 60x - 0.0004x^2\). This kind of manipulation simplifies complex expressions, paving the way for solving the equation using methods like the quadratic formula.
The approach facilitates finding quantities, optimizes calculations, and enhances understanding of mathematical relationships in economic contexts.