Polynomial Inequalities in Break-Even Analysis
We can use polynomial inequalities to determine when companies will be profitable by comparing revenue and cost functions. While real business models involve many variables, simplified polynomial models help illustrate the mathematical relationship between production levels and profitability.
Break-Even Analysis
When a company’s revenue and cost functions are both polynomials, finding when the business is profitable requires solving a polynomial inequality.
A small manufacturing company has:
- Revenue function: [latex]R(x) = 12x - 0.02x^2[/latex] (in thousands of dollars)
- Cost function: [latex]C(x) = 2x + 100[/latex] (in thousands of dollars)
where [latex]x[/latex] represents units produced (in hundreds).
To find when the company is profitable, we solve [latex]R(x) > C(x)[/latex]: [latex]12x - 0.02x^2 > 2x + 100[/latex]
Rearranging: [latex]-0.02x^2 + 10x - 100 > 0[/latex]
Multiplying by -50 (and flipping the inequality): [latex]x^2 - 500x + 5000 < 0[/latex]
Using the quadratic formula: [latex]x = \frac{500 \pm \sqrt{250000 - 20000}}{2} = \frac{500 \pm \sqrt{230000}}{2}[/latex]
[latex]x = \frac{500 \pm 479.6}{2}[/latex], so [latex]x ≈ 10.2[/latex] or [latex]x ≈ 489.8[/latex]
Since the coefficient of [latex]x^2[/latex] is positive, the parabola opens upward, making the expression negative between the roots.
The company is profitable when producing between 1,020 and 48,980 units.
Complex Cost Structures
Some businesses have more complex cost structures that create polynomial models, requiring careful analysis of multiple break-even points.
How To: Solving Polynomial Inequalities for Profitability
- Set up the inequality [latex]R(x) > C(x)[/latex] for profitability
- Move all terms to one side: [latex]R(x) - C(x) > 0[/latex]
- Factor the resulting polynomial or find its zeros
- Test intervals between zeros to determine where the expression is positive
- Consider practical constraints (production levels must be non-negative)
- Interpret the solution in business terms
A tech startup has:
- Revenue: [latex]R(x) = 15x[/latex] (thousands of dollars)
- Costs: [latex]C(x) = x^3 - 9x^2 + 26x + 24[/latex] (thousands of dollars)
where [latex]x[/latex] represents months of operation.
For profitability: [latex]15x > x^3 - 9x^2 + 26x + 24[/latex]
Rearranging: [latex]0 > x^3 - 9x^2 + 11x + 24[/latex] Or: [latex]x^3 - 9x^2 + 11x + 24 < 0[/latex]
To find zeros of [latex]f(x) = x^3 - 9x^2 + 11x + 24[/latex], we test rational roots.
Testing [latex]x = -1[/latex]: latex^3 – 9(-1)^2 + 11(-1) + 24 = -1 – 9 – 11 + 24 = 3 \neq 0[/latex] Testing [latex]x = -2[/latex]: latex^3 – 9(-2)^2 + 11(-2) + 24 = -8 – 36 – 22 + 24 = -42 \neq 0[/latex] Testing [latex]x = 3[/latex]: [latex]27 - 81 + 33 + 24 = 3 \neq 0[/latex] Testing [latex]x = 4[/latex]: [latex]64 - 144 + 44 + 24 = -12 \neq 0[/latex]
Let’s try [latex]x = 8[/latex]: [latex]512 - 576 + 88 + 24 = 48 \neq 0[/latex] Try [latex]x = 6[/latex]: [latex]216 - 324 + 66 + 24 = -18 \neq 0[/latex]
Using numerical methods, suppose we find the roots are approximately [latex]x = -1.5[/latex], [latex]x = 2.8[/latex], and [latex]x = 7.7[/latex].
Testing intervals for [latex]x ≥ 0[/latex] (since negative months don’t make sense):
- For [latex]0 < x < 2.8[/latex]: Test [latex]x = 1[/latex]: [latex]1 - 9 + 11 + 24 = 27 > 0[/latex]
- For [latex]2.8 < x < 7.7[/latex]: Test [latex]x = 5[/latex]: [latex]125 - 225 + 55 + 24 = -21 < 0[/latex] ✓
- For [latex]x > 7.7[/latex]: Test [latex]x = 9[/latex]: [latex]729 - 729 + 99 + 24 = 123 > 0[/latex]
The startup is profitable approximately between month3 and month 8.