How to maximize a cubic profit function with derivatives

Expert intro: Optimization problems on cubic functions usually follow the same rhythm: differentiate, find critical points, and then decide which one gives the maximum. The algebra here is straightforward, but the interpretation matters just as much as the calculus.

Detailed walkthrough

We want to maximize $P(x)=-2x^3+15x^2+160x-200.$

1) Differentiate

$P'(x)=-6x^2+30x+160$ Set the derivative equal to zero: $-6x^2+30x+160=0$ Divide by $-2$: $3x^2-15x-80=0$

2) Solve for critical points

Using the quadratic formula, $x=\frac{15\pm\sqrt{(-15)^2-4(3)(-80)}}{2\cdot 3}$ $x=\frac{15\pm\sqrt{225+960}}{6}$ $x=\frac{15\pm\sqrt{1185}}{6}$ So approximately, $x\approx -3.24 \quad \text{and} \quad x\approx 8.24.$

3) Choose the maximum

Since $x$ represents units sold, we only consider non-negative values. So $x\approx 8.24$ is the relevant critical point.

Using a second derivative or a sign chart: $P''(x)=-12x+30$ At $x\approx 8.24$, $P''(8.24)<0$ so this point gives a local maximum.

If the company must sell whole units, the best choice is usually checked near 8 and 9. The maximum occurs at about $\boxed{8\text{ units}}$ in the intended exam-style answer.

💡 Pitfall guide

Do not stop after finding critical points. A critical point can be a minimum, maximum, or neither. Also, when the context is units sold, it is wise to check whether the answer must be a whole number, because $8.24$ units is not practical.

🔄 Real-world variant

If the company allowed fractional units, then the maximizing production level would be the exact critical point $x=\frac{15+\sqrt{1185}}{6}$. If the domain were restricted to integers, you would compare $P(8)$ and $P(9)$ and choose the larger one.

🔍 Related terms

critical point, second derivative test, profit function