Question

Production Level 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).

Short answer

The revenue equals cost, or breaking even, at production level \(x=0\). This could mean, that if there's no production, then there's no cost nor revenue.

Step-by-step solution

Understand the Revenue and Cost Functions

Given two functions, the revenue function denoted as r(x), is \(6x\). The cost function denoted as c(x), is \(x^{3}-6 x^{2}+15 x\).

Equate Revenue and Cost Functions

To find out when both of these functions are equal, which is when the company breaks even, we set the two functions equal to each other: \(6x = x^{3}-6 x^{2}+15 x\).

Simplify the Equation

Simplify the equation by subtracting \(6x\) from both sides and collect like terms: \(0 = x^{3}-6 x^{2}+15 x - 6x\), which results in \(0 = x^{2}(x-6) + 9x\).

Solve for x

Now solve for 'x'. Since both terms include 'x', it's possible to be factored out: \(0 = x(x - 6 + 9)\). This equation equals to zero only if \(x = 0\) or \((x - 6 + 9) = 0\). Solving \((x - 6 + 9) = 0\) gives \(x = -3\).

Analyze the Results

In this context, negative production levels \(-3\) are meaningless and can be disregarded. However, break-even point at production level \(0\) makes sense, which means that the company makes neither a profit nor a loss. Any production level besides zero will yield a loss, therefore, best performance is to break even at \(x = 0\).

Key concepts

Understanding Revenue and Cost Functions

Revenue and cost functions are foundational concepts in economics and business, especially in the context of determining the financial health of a company. The revenue function, denoted here as r(x), expresses how revenue varies with different levels of production or sales. For instance, if r(x) = 6x, it means that for every unit produced and sold, the revenue increases by 6 units of currency. In real-world scenarios, this could represent selling items at a constant price without discounts or bulk pricing variations.

On the other hand, the cost function, c(x), showcases the relationship between costs and production levels. With a cost function of c(x) = x³ - 6x² + 15x, it's clear the costs are not linear but increase at a changing rate as production x increases. This could account for the variable costs of materials and labor, as well as fixed costs that do not change with production levels.

When you’re dealing with these functions, you're essentially capturing how the inflow of revenue could balance the outflow of costs — a situation known as breaking even. This balance point is where a business doesn't make a profit, but also doesn't incur a loss.

Solving Equations for Break-Even Analysis

Simplifying the Equation

The first step in solving equations for break-even analysis is setting the revenue and cost functions equal to each other because, at break-even, they are the same: Revenue = Cost. So, from our example, 6x = x³ - 6x² + 15x. To solve this, one would subtract 6x from both sides to consolidate x-terms which simplifies the equation.

Isolating the Variable

After simplification, you want to get everything on one side to set the equation to zero, making it easier to identify the values of x that satisfy the condition. The ultimate goal is to isolate x and find the possible solutions for the production level that would lead to breaking even.

Understanding the Results

Once you've found the solution for x, it is crucial to analyze whether the results make practical sense in a real-world context. In our scenario, a negative production level such as x = -3 is not feasible, as you cannot produce negative units, so such solutions are disregarded. What remains is logically deducing that a zero production level means no production, no sales, and therefore no profit or loss.

Factoring Polynomials in Break-Even Analysis

Factoring polynomials is a powerful algebraic tool often used when dealing with cost and revenue functions in break-even analysis. It simplifies complex equations into more manageable forms, often revealing the solutions within.

Remember that a polynomial, such as the cost function c(x) = x³ - 6x² + 15x, is an expression consisting of variables and coefficients, involving terms in the form of powers of x. Here, the simplification process includes factoring common factors out. For our given solution, x can be factored out, since each term includes x, this leads to 0 = x(x² - 6x + 9).

Knowing how to factor such equations is essential for understanding break-even analysis since factoring out the variable x narrows down potential break-even points. Once factored, finding the roots of the equation can unveil the production levels where revenue and cost intersect, signaling the break-even points. However, these solutions should always be examined for their real-world applicability, as some mathematical solutions might not carry practical business significance.