Question:
If $V$ is the volume of a cuboid of dimensions $a, b, c$ and $S$ is its surface area then prove that $\frac{1}{V}=\frac{2}{S}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$.
Solution:
Let the length, breadth and height of the cuboid be a, b and c, respectively.
∴ Surface area of the cuboid, S = 2(ab + bc + ca)
Volume of the cuboid, V = abc
Now,
$\frac{S}{V}=\frac{2(a b+b c+c a)}{a b c}$
$\Rightarrow \frac{S}{V}=2\left(\frac{a b}{a b c}+\frac{b c}{a b c}+\frac{c a}{a b c}\right)$
$\Rightarrow \frac{S}{V}=2\left(\frac{1}{c}+\frac{1}{a}+\frac{1}{b}\right)$
$\Rightarrow \frac{1}{V}=\frac{2}{S}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$