Question

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

Answer 1

Let Length be $x$, Breadth be $x$ ,Height be $y$ of a closed cuboid with a square base .

Given Volume $V$ i.e. Volume is constant

$V=x.x.y$

$V=x^{2}y$

$y=\frac{V}{x^2}$

Let $S$ be the surface area of closed cuboid.

$S=2x^2+4xy$

$S=2x^2+4x\frac{V}{x^2}$

$S=2x^2+\frac{4V}{x}$

$\frac{ds}{dx}=4x-\frac{4V}{x}$

For maxima and minima$\frac{ds}{dx}=4x-\frac{4V}{x}=0$

$V=x^3$

$x=V^{\frac{1}{3}}$

$y=\frac{V}{x^2}=V^{\frac{1}{3}}=x$

$\frac{d^{2}s}{d{t}^2}=4+\frac{8V}{x^3}$

At $V=x^3$

$\frac{d^{2}s}{d{t}^2}=12$

$\frac{d^{2}s}{d{t}^2}>0$

Thus minima exist,when

$Length=breadth=Height$

:- Cuboid is a cube