Question

A closed cylinder has volume $2156{cm}^3$. What will be the radius of its base so that its total surface area is minimum?

Answer 1

$Letrandhberadiusandheight.$

$Volumeofcylinder=\pi r^{2}h$

$Here\pi r^{2}h=2156$

$h=\frac{2156}{\pi r^2}⟹\left(1\right)$

$LettotalsurfaceareaofcylinderbeS.$

$S=(2\pi r^2+2\pi rh)$

$=(2\pi r^2+2\pi r\frac{2156}{\pi r^2})$

$=(2\pi r^2+\frac{4312}{r})$

$Diffwrtor,$

$\frac{ds}{dr}=(4\pi r-\frac{4312}{r^2})$

$\frac{d^{2}s}{d{r}^2}=(4\pi +\frac{8624}{r^3})$

$Now\frac{ds}{dr}=0$

$(4\pi r-\frac{4312}{r^2})=0$

$\frac{4\pi r^3-4312}{r^2}=0$

$4\pi r^3=4312$

$r^3=\frac{4312}{4\pi }$

$r=\sqrt[3]{3\frac{4312\ast 7}{4\ast 22}}$

$r=\sqrt[3]{3343}$

$r=7$