Question

If V is the volume of a cuboid of dimensions a, b, c and S is its surface area then prove that $\frac{1}{V}=\frac{2}{S}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$.

Answer 1

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,

$$\begin{gathered} \frac{S}{V}=\frac{2\left(ab+bc+ca\right)}{abc} \\ \Rightarrow \frac{{\displaystyle S}}{{\displaystyle V}}=2\left(\frac{ab}{abc}+\frac{{\displaystyle bc}}{{\displaystyle abc}}+\frac{{\displaystyle ca}}{{\displaystyle abc}}\right) \\ \Rightarrow \frac{{\displaystyle S}}{{\displaystyle V}}=2\left(\frac{1}{c}+\frac{{\displaystyle 1}}{{\displaystyle a}}+\frac{{\displaystyle 1}}{{\displaystyle b}}\right) \\ \Rightarrow \frac{1}{V}=\frac{2}{S}\left(\frac{1}{a}+\frac{{\displaystyle 1}}{{\displaystyle b}}+\frac{{\displaystyle 1}}{{\displaystyle c}}\right) \end{gathered}$$