A box with a square base and open top must have a volume of $4000c{m}^3$.

Find the dimensions of the box that minimize the amount of material used.
Find:

Sides of base in cm
Height in cm

To calculate dimensions of the open box, having volume $=4000c{m}^3$which minimize the material used.

Step-1: Find the equation to be minimized

Volume = $4000$

Box's material comprises of bottom and the sides only (no top)

Letting the bottom square of each side = $\mathrm{X}$, and the height = $\mathrm{H}$

we have 1 bottom (${\mathrm{X}}^2$) area, and $4$ sides of dimensional = $\mathrm{XH}$

so the material ($\mathrm{M}$) we want to minimize is then

$\mathrm{M}={\mathrm{X}}^2+4\mathrm{XH}$ {Equation to be minimized}

Step-2: Substitute values to find the volume in terms of X.

$\begin{array}{rcl}Volume& =& (\mathrm{Bottom}\mathrm{area}x\mathrm{height})\\ 4,000& =& X^{2}H\end{array}$

Put $\mathrm{H}$ in terms of $\mathrm{X}$ from the volume equation, or

$\mathrm{H}=\frac{4,000}{{\mathrm{X}}^2}$

Substitute for $H$ in Material equation, to be minimized:

$\begin{array}{rcl}\mathrm{M}& =& {\mathrm{X}}^2+4X·\frac{4,000}{{\mathrm{X}}^2}\\ & \Rightarrow & \mathrm{M}={\mathrm{X}}^2+\frac{16,000}{X}\\ & \Rightarrow & \mathrm{M}={\mathrm{X}}^2+16,000·{\mathrm{X}}^{-1}\end{array}$

Step-3: Find the derivative.

Take derivative and equate it to zero to minimize:

$\begin{array}{rcl}\frac{DM}{DX}& =& 2\mathrm{X}-16,000{\mathrm{X}}^{-2}\\ & =& 0\end{array}$

And,

$\begin{array}{rcl}\frac{2\mathrm{X}\left({\displaystyle \frac{{\mathrm{X}}^2}{{\mathrm{X}}^2}}\right)-16,000}{{\mathrm{X}}^2}& =& 0\\ & \Rightarrow & \frac{2{\mathrm{X}}^3-16,000}{{\mathrm{X}}^2}=0\\ & \Rightarrow & 2\frac{{\mathrm{X}}^3-8,000}{{\mathrm{X}}^2}=0\end{array}$

Step 4. Solve for X and calculate the value.

So, ${\mathrm{X}}^3-8000=0$

$\begin{array}{rcl}{\mathrm{X}}^3-8000& =& 0\\ & \Rightarrow & {\mathrm{X}}^3=8,000\\ & \Rightarrow & \mathrm{X}=20\end{array}$

The Volume $\left({\mathrm{X}}^2\mathrm{H}\right)$ is then $400\mathrm{H}=4,000$ , So $\mathrm{H}=10$.

Hence, the dimensions are $20\mathrm{x}20\mathrm{x}10$which minimize the material, of an open cuboid box with a given volume.