sending current in an external resistance $\mathrm{R}$. Then current in the external circuit be i, then $i=\frac{m n E}{n r+m R}$

It is clear from this equation, that for current to be maximum the value of $(\mathrm{nr}+\mathrm{mR})$ shouid be minimum ie, $n r+m R=[\sqrt{n r}-\sqrt{m R}]^{2}+2 \sqrt{m n R r}$

It's minimum value is zero.

ie, $[\sqrt{n r}-\sqrt{m R}]^{2}=0$

$\sqrt{n r}-\sqrt{m R}=0$

$n r=m R$

$\Rightarrow R=\frac{n r}{m}$

but $\frac{n r}{m}$ is the internal resistance of the whole battery. Thus, current is maximum when the internal resistance of battery is equal to external resistance.