Problem 75 - College Algebra and Calculus: An Applied Approach

Question

The demand equation for a product is \(p=50-0.0005 x\) 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 \(\$ 250,000 ?\)

Step-by-Step Solution

Answer

The number of units that must be sold to generate a revenue of $250,000 is approximately 625,000.

Step 1: Write Down Given Equations

The given equations are R = xp and p = 50 - 0.0005x. We are asked to find the x value, which brings R equals to $250,000.

Step 2: Substitute for p in Revenue Equation

Substitute value of p (from p = 50 - 0.0005x) into the revenue equation. The new equation is \(R = x(50 - 0.0005x)\)

Step 3: Input the Known Revenue

We know that R equals $250,000. Substitute this into the equation. The revised equation is \(250000 = x(50 - 0.0005x)\)

Step 4: Solve the Equation

Solve for x in the equation. This leads to a quadratic equation \(0.0005x^2 - 50x + 250000 = 0\). Solving the quadratic equation either by using quadratic formula or other suitable methods, you get two roots. Choose the positive root as the number of units sold cannot be negative.

Key Concepts

Demand Equation

The demand equation, such as the one given in the exercise, helps to express the relationship between the price of a product ( p ight) and the quantity sold ( x ight). It is typically written in a linear form:

• p = a - b imes x

This formula helps businesses understand how the price affects the demand; a decrease in price might increase the number of units sold. It also reveals consumer behavior and preferences by showing how price and quantity demanded vary. One practical application is in setting the right price point for maximizing sales revenue without discouraging customers with high prices. Understanding the demand equation is crucial for companies when planning marketing strategies and forecasting sales.

Quadratic Equation

Quadratic equations are essential when dealing with complex revenue calculations, especially when those involve squared terms.The general form is:

• ax^2 + bx + c = 0

In our context, solving the revenue equation involves setting the quadratic equation formed by substituting the demand function into the revenue function as zero.Quadratic equations can have up to two solutions, calculated using the quadratic formula:

• x = \( \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

These solutions represent possible values for the number of units sold. However, only the positive, realistic value makes sense in a sales context. Understanding how to solve quadratic equations is a key skill in this type of revenue calculation problem.

Unit Price

In any demand equation, the unit price is what p ight represents, derived from the relationship between price and sales quantity. It shows what customers are willing to pay as more units are sold. In our example, it is expressed in terms of the quantity, x ight: p = 50 - 0.0005x . Unit pricing is crucial for both pricing strategy and revenue optimization. By understanding how lower prices could potentially boost sales and revenue, businesses can set competitive prices. Moreover, recognizing the relationship between unit price and demand guides decisions about production and inventory levels.

Product Sales

Product sales, represented by x ight in both the demand and revenue equations, highlight the volume of sales needed to achieve certain financial targets. In the revenue equation, R = x imes p , the focus is to determine how many units must be sold to reach a target revenue. For example, achieving a revenue of $250,000 requires determining the correct number of unit sales, using both the demand equation and solving a quadratic equation. Successfully calculating product sales aids businesses in setting realistic sales goals, planning production, and even determining marketing budgets to meet desired revenue outcomes.