Show that the surface area of a closed cuboid with square base and given volume is minimum, when it is a cube.

Let Length be x, Breadth be x, Height be y of a closed cuboid with a square base.
$\begin{array}{l}\mathrm{V}=\mathrm{x}\cdot \mathrm{x}\cdot \mathrm{y}\\ \mathrm{V}={\mathrm{x}}^2\mathrm{y}\\ \mathrm{y}=\frac{\mathrm{V}}{{\mathrm{x}}^2}\end{array}$
Given Volume V i.e. Volume is constant Let S be the surface area of closed cuboid.
Now surface area of cuboid
$\begin{array}{l}\mathrm{S}=2{\mathrm{x}}^2+4\mathrm{xy}\\ \mathrm{S}=2{\mathrm{x}}^2+4\mathrm{x}\frac{\mathrm{V}}{{\mathrm{x}}^2}\\ \mathrm{S}=2{\mathrm{x}}^2+\frac{4\mathrm{V}}{\mathrm{x}}\\ \frac{\mathrm{ds}}{\mathrm{dx}}=4\mathrm{x}-\frac{4\mathrm{V}}{{\mathrm{x}}^2}\end{array}$

$$\begin{gathered} for maxima and minima \frac{dS}{\mathrm{dx}}=4\mathrm{x}-\frac{4\mathrm{V}}{x^{2_{ }}}=0 \\ \Rightarrow 4x^3=4V \\ \Rightarrow V=x^3 \end{gathered}$$
Thus, minima exist, when
Length = breadth = Height
$\therefore$Cuboid is a cube.