A capacitor of capacitance C is connected to a battery of emf E at t = 0 through a resistance R. Find the maximum rate at which energy is stored in the capacitor. When does the rate has this maximum value?

$\begin{array}{l}\mathrm{U}=\frac{{\mathrm{q}}^2}{2\mathrm{c}}\Rightarrow \mathrm{U}=\frac{{\mathrm{q}}_0^2}{2\mathrm{c}}{\left(1-{\mathrm{e}}^{-\mathrm{t}/\mathrm{CR}}\right)}^2\\ \mathrm{R}=\frac{\mathrm{du}}{\mathrm{dt}}=\frac{{\mathrm{q}}_{\mathrm{o}}^2}{2\mathrm{c}}\left(2\left(1-{\mathrm{e}}^{-\mathrm{t}/\mathrm{CR}}\right){\mathrm{e}}^{-\mathrm{t}/\mathrm{CR}}\frac{1}{\mathrm{CR}}\right)\\ \Rightarrow \mathrm{R}=\frac{{\mathrm{q}}_0^2}{{\mathrm{C}}^2\mathrm{R}}\left({\mathrm{e}}^{-\mathrm{t}/\mathrm{CR}}-\mathrm{e}\frac{-2\mathrm{t}}{\mathrm{CR}}\right)\end{array}$
$\frac{\mathrm{dR}}{\mathrm{dt}}=0$then R will be maximum
$\begin{array}{l}\frac{{\mathrm{q}}_0^2}{{\mathrm{C}}^2\mathrm{R}}\left({\mathrm{e}}^{-\mathrm{t}/\mathrm{CR}}\left(\frac{-1}{\mathrm{CR}}\right)-{\mathrm{d}}^{\frac{-2\mathrm{t}}{\mathrm{CR}}}\left(\frac{-2}{\mathrm{CR}}\right)\right)=0\\ \Rightarrow {\mathrm{e}}^{-\mathrm{t}/\mathrm{CR}}=2\cdot {\mathrm{e}}^{\frac{-2\mathrm{t}}{\mathrm{CR}}\Rightarrow {\mathrm{e}}^{\mathrm{tCR}}=2\Rightarrow \mathrm{t}=\mathrm{CRln}2}\\ {\mathrm{R}}_{\text{maximum }}=\frac{{\mathrm{q}}_0^2}{{\mathrm{C}}^2\mathrm{R}}\left(\mathrm{e}\frac{-\mathrm{CRln}2}{\mathrm{CR}}-{\mathrm{e}}^{\frac{-2\mathrm{CRln}2}{\mathrm{CR}}}\right)\\ =\frac{{\mathrm{q}}_0^2}{{\mathrm{C}}^2\mathrm{R}}\left(\frac{1}{2}-\frac{1}{4}\right)=\frac{{\mathrm{q}}_0^2}{4{\mathrm{c}}^2\mathrm{R}}=\frac{{\mathrm{C}}^2{\mathrm{E}}^2}{4{\mathrm{C}}^2\mathrm{R}}=\frac{{\mathrm{E}}^2}{4\mathrm{R}}\end{array}$