A cell of internal resistance r drives a current through an external resistance R. The power delivered by the cell to the external resistance is maximum when

let E = e.m.f. of cell and r = its internal resistance. Current through the cell 
$\mathrm{i}=\frac{\mathrm{E}}{\mathrm{R}+\mathrm{r}}$
P = Power delivered to .R
$={\mathrm{i}}^2\mathrm{R}=\frac{{\mathrm{E}}^2\mathrm{R}}{(\mathrm{R}+\mathrm{r}{)}^2}$
Power delivered will be maximum, when
$\frac{\mathrm{dP}}{\mathrm{dR}}={\mathrm{E}}^2\left[\frac{1}{(\mathrm{R}+\mathrm{r}{)}^2}-\frac{2\mathrm{R}}{(\mathrm{R}+\mathrm{r}{)}^3}\right]=0$
Solving, we get E = r