Problem 59
Question
Show that if \(r(x)=6 x\) and \(c(x)=x^{3}-6 x^{2}+15 x\) are your revenue and cost functions, then the best you can do is break even (have revenue equal cost).
Step-by-Step Solution
Verified Answer
Break-even occurs when producing 3 units (\(x = 3\)).
1Step 1: Set Revenue Equal to Cost
To find out when profit is zero, we set the revenue function equal to the cost function. This gives us the breach even point equation: \( r(x) = c(x) \), or \( 6x = x^3 - 6x^2 + 15x \).
2Step 2: Simplify the Equation
Subtract \(6x\) from both sides to simplify the equation: \( x^3 - 6x^2 + 15x - 6x = 0 \), which simplifies to \( x^3 - 6x^2 + 9x = 0 \).
3Step 3: Factor the Equation
Factor out \(x\) from the equation: \( x(x^2 - 6x + 9) = 0 \). This can be further factored as \( x(x-3)^2 = 0 \).
4Step 4: Solve for x
Set each factor equal to zero and solve: \( x = 0 \) or \( (x-3)^2 = 0 \). The latter gives \( x = 3 \).
5Step 5: Interpret the Results
The solutions \( x = 0 \) and \( x = 3 \) indicate where the revenue equals the cost (break-even points). However, producing zero units is not practical, so production should occur when \( x = 3 \).
Key Concepts
Revenue FunctionCost FunctionFactoring PolynomialsSolving Equations
Revenue Function
The revenue function is key to understanding how much money a business makes from selling goods or services. It shows the relationship between the number of units sold and the total income generated. In our example, the revenue function is given by \( r(x) = 6x \). This means that for every unit sold, $6 is earned.
The revenue function is linear, which indicates that each additional unit sold brings in a constant amount of revenue. It's crucial for businesses to comprehend their revenue functions to make informed decisions about pricing and production. Analyzing the revenue function helps to predict how changes in sales volume can influence total revenue.
Understanding the components of a revenue function allows businesses to identify trends, set objectives, and develop strategies for growth. Overall, it provides a straightforward way to foresee the income potential across different sales levels.
The revenue function is linear, which indicates that each additional unit sold brings in a constant amount of revenue. It's crucial for businesses to comprehend their revenue functions to make informed decisions about pricing and production. Analyzing the revenue function helps to predict how changes in sales volume can influence total revenue.
Understanding the components of a revenue function allows businesses to identify trends, set objectives, and develop strategies for growth. Overall, it provides a straightforward way to foresee the income potential across different sales levels.
Cost Function
The cost function represents the total cost of production, including all expenses associated with manufacturing goods or providing services. This function typically accounts for both fixed and variable costs, showing how costs change with variations in production.
In the case of our exercise, the cost function is \( c(x) = x^3 - 6x^2 + 15x \). Here, costs scale with the cube of the product quantity as well as other polynomial terms. This form indicates that costs increase at a non-linear rate as production grows.
The cost function is especially important for budgeting, planning, and making decisions regarding the scale of production. By understanding the detailed breakdown of costs, businesses can identify areas where efficiency improvements can be made, potentially reducing costs and increasing profitability.
In the case of our exercise, the cost function is \( c(x) = x^3 - 6x^2 + 15x \). Here, costs scale with the cube of the product quantity as well as other polynomial terms. This form indicates that costs increase at a non-linear rate as production grows.
The cost function is especially important for budgeting, planning, and making decisions regarding the scale of production. By understanding the detailed breakdown of costs, businesses can identify areas where efficiency improvements can be made, potentially reducing costs and increasing profitability.
Factoring Polynomials
Factoring polynomials is a fundamental process in algebra that involves breaking down a complex expression into simpler components that can be multiplied together to get the original polynomial. It's an essential skill for solving polynomial equations and finding their roots.
In the given problem, once the revenue has been set equal to the cost, we have the equation \( x^3 - 6x^2 + 9x = 0 \). To solve this, we factor the polynomial. First, we factor out the greatest common factor, which is \( x \), resulting in \( x(x^2 - 6x + 9) = 0 \).
Further factoring involves recognizing that \( x^2 - 6x + 9 \) can be expressed as \( (x-3)^2 \), transforming our expression to \( x(x-3)^2 = 0 \). By factoring polynomials, we simplify solving equations and can more easily find where the functions reach their break-even points.
In the given problem, once the revenue has been set equal to the cost, we have the equation \( x^3 - 6x^2 + 9x = 0 \). To solve this, we factor the polynomial. First, we factor out the greatest common factor, which is \( x \), resulting in \( x(x^2 - 6x + 9) = 0 \).
Further factoring involves recognizing that \( x^2 - 6x + 9 \) can be expressed as \( (x-3)^2 \), transforming our expression to \( x(x-3)^2 = 0 \). By factoring polynomials, we simplify solving equations and can more easily find where the functions reach their break-even points.
Solving Equations
Solving equations involves finding the value(s) of variable(s) that make the equation true. It's a critical skill in mathematics, especially in analyzing financial functions such as revenue and cost.
In our break-even analysis, after factoring the polynomial \( x(x-3)^2 = 0 \), we solve for \( x \) by setting each factor equal to zero. This gives two potential solutions: \( x = 0 \) and \( x - 3 = 0 \) (which simplifies to \( x = 3 \)).
These solutions point to the values where the revenue equals the cost, known as the break-even points. However, only values that make practical sense should be considered in real-world applications. Thus, \( x = 3 \) is the meaningful solution as it represents a feasible production level, indicating that producing 3 units allows the business to break even.
In our break-even analysis, after factoring the polynomial \( x(x-3)^2 = 0 \), we solve for \( x \) by setting each factor equal to zero. This gives two potential solutions: \( x = 0 \) and \( x - 3 = 0 \) (which simplifies to \( x = 3 \)).
These solutions point to the values where the revenue equals the cost, known as the break-even points. However, only values that make practical sense should be considered in real-world applications. Thus, \( x = 3 \) is the meaningful solution as it represents a feasible production level, indicating that producing 3 units allows the business to break even.
Other exercises in this chapter
Problem 58
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