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Chapter 9: Problem 522

For the following exercises, write a system of equations to solve each problem. Solve the system of equations. A factory has a cost of production \(C(x)=150 x+15,000\) and a revenue function \(R(x)=200 x\) . What is the break-even point?

Short Answer

Expert verified
The break-even point is at 300 units.

Step by step solution

01

Understand the Problem

To find the break-even point, we need to determine the quantity where costs equal revenue. This is the point where the factory does not make a profit or a loss.
02

Write the System of Equations

We are given the cost function \(C(x) = 150x + 15,000\) and the revenue function \(R(x) = 200x\). To find the break-even point, we set these two functions equal to one another: \(150x + 15,000 = 200x\).
03

Rearrange the Equation

Rearrange the equation \(150x + 15,000 = 200x\) by subtracting \(150x\) from both sides to isolate \(x\) on one side: \(15,000 = 200x - 150x\). This simplifies to \(15,000 = 50x\).
04

Solve for x

Now, solve for \(x\) by dividing both sides by 50: \(x = \frac{15,000}{50}\). When you perform the division, you find that \(x = 300\).
05

Interpret the Solution

The break-even point occurs when \(x = 300\). This means the factory needs to produce and sell 300 units to cover its production costs and achieve no loss or gain.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Break-Even Point
The break-even point in any business scenario is the moment when total costs and total revenue are equal. In simpler terms, it's the stage where a business hasn't made any profit, nor has it incurred any losses. Think of it as the magic balance between expenses and income.
The break-even point is crucial because:
  • It helps businesses understand the minimum performance required to avoid losses.
  • Provides a clear target for businesses, aiding in strategic planning and decision making.
  • Acts as a baseline to evaluate profitability. Once this point is surpassed, any additional units will contribute to profit.
In our example, the factory's break-even point comes about by mapping when the cost function and the revenue function intersect. This intersection signifies a neutral profit line, where the production costs are precisely covered by the revenues generated.
Cost Function
The cost function represents the total cost of producing a specific number of goods. In mathematical terms, it captures both fixed and variable costs.
In our exercise, the cost function given is:
\[ C(x) = 150x + 15,000 \]
  • The term \(150x\) corresponds to the variable cost, which varies with the number of units produced, with \(150\) being the cost per unit.
  • The constant \(15,000\) indicates the fixed costs. These are costs that remain constant regardless of the production volume, such as rent, machinery costs, and utility expenses.
Understanding the cost function helps businesses to:
  • Project future expenses based on different production levels.
  • Fine-tune pricing strategies to ensure costs are covered and profits are maximized.
  • Make informed decisions about scaling production either upwards or downwards based on cost structures.
Revenue Function
A revenue function is designed to predict the amount of money generated from selling products or services. It essentially reflects the income a company expects based on the quantity of goods sold.
In the problem, the revenue function is expressed as:
\[ R(x) = 200x \]
  • This means each unit produced and sold brings in \(200\) in revenue.

    • Seeing potential revenue through this function helps companies to:
    • Estimate income based on sales projections, assisting in future planning and growth estimations.
    • Calculate the returns from production once costs are deducted.
    • Provide a framework for evaluating market performance and making pricing adjustments.
    By comparing this with the cost function, businesses can strategize their production levels and marketing efforts to achieve their target break-even and profitability thresholds.
    Solve Equations
    Solving equations is a process that involves finding the values of variables that satisfy given mathematical statements. When you're tasked with finding a break-even point, you're often solving for the point where two functions equal each other.
    In the context of our exercise, we set the cost and revenue functions equal:\[ 150x + 15,000 = 200x \]This requires eliminating one of the variables to isolate \(x\). Subtracting \(150x\) from both sides, the equation simplifies to:
    \[ 15,000 = 50x \]
    This reveals a simpler equation with only one unknown. Divide both sides by 50 to solve for \(x\):
    • \(x = \frac{15,000}{50} = 300\)
    Here, solving equations helps us to:
    • Determine specific values like the break-even quantity in a production strategy.
    • Understand relationships between costs, revenue, and output levels.
    • Use algebra to make complex business decisions with a mathematical backing.

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    Most popular questions from this chapter

    For the following exercises, solve using a system of linear equations. A small fair charges \(1.50 for students, \)1 for children, and \(2 for adults. In one day, three times as many children as adults attended. A total of 800 tickets were sold for a total revenue of \)1,050. How many of each type of ticket was sold?

    For the following exercises, decompose into partial fractions. $$ \frac{x^{3}-4 x^{2}+3 x+11}{\left(x^{2}-2\right)^{2}} $$

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    For the following exercises, perform the requested operations on the given matrices. $$A=\left[\begin{array}{cc}{4} & {-2} \\ {1} & {3}\end{array}\right], B=\left[\begin{array}{rrr}{6} & {7} & {-3} \\ {11} & {-2} & {4}\end{array}\right], C=\left[\begin{array}{cc}{6} & {7} \\ {11} & {-2} \\ {14} & {0}\end{array}\right], D=\left[\begin{array}{rrr}{1} & {-4} & {9} \\ {10} & {5} & {-7} \\ {2} & {8} & {5}\end{array}\right], E=\left[\begin{array}{rrr}{7} & {-14} & {3} \\ {2} & {-1} & {3} \\ {0} & {1} & {9}\end{array}\right]$$ $$ A^{3} $$
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