Problem 47 - Thomas Calculus

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

Answer

Breaking even occurs at \(x = 0\) and \(x = 3\).

Step 1: Find the Break-Even Point

To determine if it's possible to break even, we need to solve for when revenue equals cost. Set the revenue function equal to the cost function: \[6x = x^3 - 6x^2 + 15x\]

Step 2: Simplify the Equation

Rearrange the equation by moving all terms to one side to form a polynomial equation:\[x^3 - 6x^2 + 15x - 6x = 0\]Simplify by combining like terms:\[x^3 - 6x^2 + 9x = 0\]

Step 3: Factor the Polynomial

Factor out the common factor, \(x\), from the polynomial:\[x(x^2 - 6x + 9) = 0\]

Step 4: Solve the Factored Equation

Solve for the roots of the equation. The equation is now written as:\[x(x - 3)^2 = 0\]This gives us the solutions: \(x = 0\) and \(x = 3\).

Step 5: Check If Solutions Break Even

Evaluate if the revenue equals the cost at these points. For \(x = 0\), both revenue and cost are zero, which means break-even is achieved. For \(x = 3\), calculate:Revenue: \(r(3) = 6 \times 3 = 18\)Cost: \(c(3) = 3^3 - 6(3)^2 + 15(3) = 27 - 54 + 45 = 18\)Thus, at \(x = 3\) revenue also equals cost.

Key Concepts

Revenue Function

The revenue function is a straightforward mathematical representation used to model how much money a company earns based on the number of units sold. In our given problem, the revenue function is expressed as a linear equation: \( r(x) = 6x \). This formula tells us that for each unit (\( x \)) sold, the company earns \(6.

• Linear Relationship: The function \( r(x) = 6x \) indicates a direct linear relationship between the quantity sold and the revenue. This means for every additional unit sold, the revenue increases by \)6.
• Understanding the Slope: In this function, 6 is the slope, which represents the rate of revenue increase. The steeper the slope, the faster the revenue grows as sales increase.

This uncomplicated formula helps us understand and predict sales outcomes quickly. Given its simplicity, determining revenue for different numbers of units sold is a breeze.

Cost Function

A cost function, like the one given in this exercise, models how much it costs a company to produce units. Our cost function is a bit more complex: \( c(x) = x^3 - 6x^2 + 15x \). This polynomial equation involves more variables, reflecting the complex nature of costs in production.

• Non-linear Costs: The cost function here is a polynomial, indicating that the cost of production doesn't increase linearly with the number of units produced. It depends on a cubic polynomial, which can have different rates of increase or decrease depending on the number of units \( x \).
• Fixed and Variable Costs: While not explicitly broken down here, polynomial cost functions usually encapsulate both fixed costs (independent of production volume) and variable costs (changing with production volume).

Such a function emphasizes the importance of understanding production complexities and how costs might behave unpredictably as output increases.

Polynomial Equations

Polynomial equations are an essential mathematical tool in analyzing and solving various real-world problems, including economic models like revenue and cost analysis. A polynomial equation consists of multiple terms, each with variables raised to whole number powers, and the variable is typically denoted as \( x \).

• Understanding Terms: In \( x^3 - 6x^2 + 9x \), each term represents different influences on the overall value based on the power and coefficient. The degree, or the highest power of \( x \), indicates the equation's complexity.
• Simplifying and Rearranging: To solve for \( x \), one must usually combine like terms, rearrange and simplify the polynomial equation. This might involve bringing all terms to one side of the equation, as done in this exercise.

Being familiar with these equations is crucial for problem-solving across various domains, especially in business and economics.

Factoring Polynomials

Factoring polynomials is a method used to break down a polynomial into simpler 'factor' expressions that can be multiplied to get back the original polynomial. In breaking down \( x^3 - 6x^2 + 9x \), the common factor \( x \) was extracted to simplify the equation into \( x(x - 3)^2 = 0 \).

• Finding Factors: The goal is to identify common terms or special patterns (like squares or cubes) that allow the polynomial to be expressed as a product of simpler expressions.
• Solving for Roots: Once factored, the equation \( x(x - 3)^2 = 0 \) is simpler to solve, yielding values of \( x \) that satisfy the equation: \( x = 0 \) and \( x = 3 \).

Factoring is instrumental in solving equations, finding zeroes or roots, and simplifying complex expressions to understand their behavior better.