Show that the height of a cylinder, which is open at the top,having a given surface area and greatest volume,is equal to the radius of its base. <div style="flo

Show that the height of a cylinder, which is open at the top,having a given surface area and greatest volume,is equal to the radius of its base.

Answers (1)

Let R be the radius
H be the height
V be the volume
s be the total surface area

$V= \pi R^{2}H$
$s= \pi R^{2}+2\pi RH$
$H= \frac{s-\pi R^{2}}{2\pi R}----(1)$
substititing value of H in V
$V= \frac{1}{2}\left ( sR-\pi R^{3} \right )\rightarrow differentiating\: equation:$
$\frac{dV}{dR}= \frac{1}{2}\left ( s-3\pi R^{2} \right )$
$\frac{dV}{dR}= 0$
$\Rightarrow \frac{1}{2}\left ( s-3\pi R^{2} \right )= 0---(2)$
$R= \sqrt{\frac{s}{3\pi }}$
$R= \sqrt{\frac{s}{3\pi }}---(3)$
Differentiating equation (2) again
$\frac{d^{2}V}{dR^{2}}= \frac{1}{2}\left ( 0-6\pi R \right )\Rightarrow -3\pi R$
V is greatest value $R= \sqrt{\frac{s}{3\pi }}$ from equation (3)
$H= \frac{s-\pi \times \frac{s}{3\pi }}{2\pi \sqrt{\frac{3}{3\pi }}}$ [ substituting equation (1) $R= \sqrt{\frac{s}{3\pi }}$ ]
$H= \frac{\frac{2s}{3}}{2\sqrt{\frac{\pi s}{3}}}= \sqrt{\frac{s}{3\pi }}=R$