Problem 75 - Algebra and Trigonometry Real Mathematics, Real People

Question

Use a graphing utility to graph the cost and revenue functions in the same viewing window. Find the sales \(x\) necessary to break even \((R=C)\) and the corresponding revenue \(R\) obtained by selling \(x\) units. (Round to the nearest whole unit.) Cost \(C=8650 x+250,000\) Revenue \(R=9950 x\)

Step-by-Step Solution

Answer

The break-even sales volume \(x\) and the corresponding revenue \(R\) can be found by graphing the cost and revenue functions and finding their intersection point or by solving the equation \(8650x + 250,000 = 9950x\) and substituting the value of \(x\) back into the revenue function.

Step 1: Graph the functions

Use a graphing utility to plot the functions for Cost \(C=8650x + 250,000\) and Revenue \(R=9950x\). Both functions should be represented on the same graph for easy visualization of the intersection point.

Step 2: Identifying the intersection point

The point where both the cost and revenue functions intersect is the breakeven point. This point is where the revenue equals the cost. Note down the x value at this point as this represents the number of units that need to be sold to break even.

Step 3: Solve for x

Using algebra, the sales volume necessary to break even can also be found by setting the Cost function equal to Revenue function, and solving for \(x\):So, \(8650x + 250,000 = 9950x\). Simplify this equation to find the break-even sales volume \(x\). Always round \(x\) to nearest whole units as units cannot be in decimals.

Step 4: Find corresponding revenue

Substitute \(x\) obtained from Step 3 into the revenue function R to get the corresponding revenue. \(R = 9950 \times x\)

Key Concepts

Cost Function

The cost function, denoted as \(C\), provides a mathematical representation of the total costs incurred by a business when producing a certain number of goods or services. In this exercise, the cost function is given by \(C = 8650x + 250,000\). This equation highlights two important components:

• Fixed Costs: These are costs that do not change with the level of production. In this case, the fixed cost is \(250,000\), which might include expenses like rent or salaries that remain constant regardless of how many units are produced.
• Variable Costs: These increase with each additional unit produced. Here, the variable cost per unit is \(8650\), which represents costs such as materials and labor that fluctuate with production levels.

The cost function helps businesses understand overall expenses associated with different production volumes, which is crucial for financial planning and analysis.

Revenue Function

The revenue function, represented by \(R\), describes how much money a company makes from selling its goods or services. In this case, the given revenue function is \(R = 9950x\), where \(x\) is the number of units sold. The function is straightforward as it is linear, indicating:

• Price per Unit: The coefficient \(9950\) represents the selling price per unit. This is the amount a company earns with each unit sold.

The simplicity of this function suggests that there are no discounts or price changes regardless of the volume of sales. Understanding the revenue function is crucial because it allows businesses to forecast potential earnings and make informed decisions.

Graphing Utilities

Using graphing utilities is an essential technique for visually analyzing and comparing mathematical functions. In this exercise, you are asked to graph the cost \(C\) and revenue \(R\) functions using these tools.

• Visual Comparison: With both functions plotted on the same set of axes, you can visually identify where they intersect. This intersection represents the break-even point.
• Ease of Understanding: Graphing shows changes and trends in a clear way. It helps you see how costs and revenues grow with each unit.

Graphing utilities often have features for inputting equations and adjusting the viewing window to properly fit the relevant details, providing an intuitive way to solve complex problems.

Algebraic Solutions

An algebraic solution is a step-by-step process involving mathematical operations to find variable values and solve equations. For the break-even analysis in this exercise, the algebraic approach involves:

• Setting the cost function equal to the revenue function: \(8650x + 250,000 = 9950x\).
• Simplifying the equation to isolate \(x\), the number of units that need to be sold to break even.
• Solving for \(x\) involves rearranging the equation to find the difference in coefficients and dividing to find the point of intersection.

This procedural approach not only determines the break-even quantity but also provides a foundation for understanding how sales and costs interplay. It's crucial to always round \(x\) to the nearest whole number in real-world applications because you can't sell a fraction of a unit.