Maximize Volume of an Open Box

Using the above diagram the volume $V = (l - 2x)^2 x = 4x^3 - 4lx + l^2 x \implies$

$\dfrac{dv}{dx} = 12x^2 - 8lx + l^2 = 0 \implies x = \dfrac{8l \pm 4l}{24} = \dfrac{l}{2}, \dfrac{l}{6}$

$x = \dfrac{l}{2} \implies V = 0 \therefore$ we drop $x = \dfrac{l}{2}$ and choose $x = \dfrac{l}{6} \implies V = \dfrac{2}{27}l^3 = 2$

$\implies l = \boxed{3}$ .

Note: Using the second derivative test we obtain:

$\dfrac{d^2V}{dx^2}|_{x = \dfrac{l}{6}} = (24x - 8l)|_{x = \dfrac{l}{6}} = -4l < 0 \implies$ local max at $x = \dfrac{l}{6}$

Let each side of each small square be of length $x$ . Then $V=x(l-2x)^2$ . Now the A.M. of $l-2x,l-2x$ and $4x$ is greater than or equal to their G.M. That is, $\dfrac{l-2x+l-2x+4x}{3}\geq {\sqrt[3] {4x(l-2x)^2}}$ or $4x(l-2x)^2\leq (\dfrac{2l}{3})^3$ , or $x(2l-x)^2\leq \dfrac{2l^3}{27}$ . Therefore $\dfrac{2l^3}{27}=2$ or $l=\boxed 3$