Differential Calculus: Cylinder of largest lateral area inscribed in a sphere

$(2r)^2 + h^2 = (2R)^2$

$4r^2 + h^2 = 4R^2$

$h = \sqrt{4R^2 - 4r^2}$
 

$S = 2\pi rh$

$S = 2\pi r\sqrt{4R^2 - 4r^2}$

$\dfrac{dS}{dr} = 2\pi \left[ r \cdot \dfrac{-8r}{2\sqrt{4R^2 - 4r^2}} + \sqrt{4R^2 - 4r^2} \right] = 0$

$\sqrt{4R^2 - 4r^2} = \dfrac{8r^2}{2\sqrt{4R^2 - 4r^2}}$

$4R^2 - 4r^2 = 4r^2$

$R^2 = 2r^2$

$r^2 = \frac{1}{2}R^2$

$r = \frac{1}{\sqrt{2}}R$
 

$h = \sqrt{4R^2 - 4r^2}$

$h = \sqrt{4R^2 - 4(\frac{1}{2}R^2)}$

$h = \sqrt{2R^2}$

$h = \sqrt{2}R$
 

$\text{Required ratio} = \dfrac{h}{r}$

$\text{Required ratio} = \dfrac{\sqrt{2}R}{\frac{1}{\sqrt{2}}R}$

$\text{Required ratio} = 2$       answer