Why sphere minimizes surface area for a given volume? I was studying surface tension recently. Rain drops or bubbles of any kind which form are always of a spherical shape. 
This is because the liquid tries to minimize the surface area as the molecules on the surface have higher potential energy. The shape which most efficiently manages to do this is a sphere. This is the part I didn't understand; how exactly do we reach the conclusion that a sphere would be the most efficient shape? Why couldn't it be any other shape?
 A: The units of surface tension are $[N/m]=[J/m^2]$ which means surface tension can be interpreted as the energy cost of creating additional surface area. Imagine any shape in equilibrium; increasing its surface area will require an energy input to overcome surface tensile forces before it reaches a new equilibrium.
Now per volume the surface area of a cube of side $s$ is $a_c = \frac{6s^2}{s^3} = \frac{6}{s}$ whilst a sphere of radius $r$ is $a_s = \frac{4\pi r^2}{\frac{4}{3}\pi r^3} = \frac{3}{r}$. So for equal volumes $s^3=\frac{4}{3}\pi r^3\rightarrow \frac{s}{r} = \sqrt[3]{\frac{4}{3}\pi}$ we find:
$$\frac{a_s}{a_c} = \frac{1}{2}\frac{s}{r} = \sqrt[3]{\frac{\pi}{6}}<1$$
which mathematically shows that the specific surface area of a sphere is less than that of a cube. In fact this can be shown for any shape:

As you can see the shape of a sphere has the lowest possible surface area to volume ratio and therefor requires the least energy to maintain its shape. The minimization of energy cost is usually what drives the physical world, hence natural objects like bubbles and raindrops tend to a spherical shape.
A: $\mathbf{Geometrical \ approach}:$ Each point on the surface of a sphere is at an equal distance (equal to radius) from the center this results in the minimum surface area for a given volume.  This can be proved analytically by comparing the surface area of a sphere with that of any other geometrical shape for a given volume.  
For example, let's compare the surface areas of a sphere & a cube for a given volume say $V$ of the drop,
For the sphere of a radius $r$ $$\frac{4\pi}{3}r^3=V\implies r=\left(\frac{3V}{4\pi}\right)^{1/3}$$
surface area of the sphere $\color{red}{S_1}=4\pi r^2=4\pi\left(\frac{3V}{4\pi}\right)^{2/3}=\color{red}{\left(\frac{9}{2\sqrt \pi}\right)^{1/3}V^{2/3}\approx1.36 V^{2/3}}$
For a cube with edge length $a$
$$a^3=V\implies a=V^{1/3}$$
surface area of the  cube $\color{blue}{S_2}=6 a^2=\color{blue}{6V^{2/3}}$
Comparing surface areas, $\color{red}{S_1}<\color{blue}{S_2}$ i.e. the surface area of a sphere is smaller than that of a cube for a given volume 
Similarly, we can analytically compare surface area of a sphere with that of any other geometrical shape. 
The minimum surface area of the sphere results in the minimum surface energy of the drop. 
 that's why the drop takes spherical shape to minimize its potential energy 
