Cut corners from rectangle to get box with max volume

Question

I've got a rectangle (no informations about the box, volume box etc.). I need to find how much should I cut from the rectangle to get a box with maximum volume, so I need to find $x = ?$

At the moment I got to: $V(x) = (a-2x)(2a-2x)x = 4x^3 - 6ax^2 + 2a^2x$ and interval $[0,a/2]$.

$V'(x) = 12x^2 - 12ax - 2a^2$

How should I find $x$ now? I'm not really sure how should I continue.

Thanks.

Answer 1

Assume $a\space\wedge\space x\in\mathbb{R}^+$:

• $$\text{V}(x)=x\left(a-2x\right)\left(2a-2x\right)$$
• $$\text{V}'(x)=2\left(a^2-6ax+6x^2\right)$$

Now solve $\text{V}'(x)=0$:

$$2\left(a^2-6ax+6x^2\right)=0\Longleftrightarrow$$ $$a^2-6ax+6x^2=0\Longleftrightarrow$$

---

Use he quadratic formula:

---

$$x=\frac{a\left(3\pm\sqrt{3}\right)}{6}$$