Show that the right circular cylinder of given surface

Question: Show that the right circular cylinder of given surface and maximum volume is such that its height is equal to the diameter of the base.

Solution: Let $r$ be the radius of the circular base and $h$ be the height of closed right circular cylinder.

Formula for Total surface area $(\mathrm{S})=2 \pi r h+2 \pi r^2$

$S-2\pi r^2 = 2\pi rh$

$\Rightarrow h= \frac{S-2\pi r^2}{2\pi r}$ –(i)

Volume of cylinder $(V) = \pi r^2 h$

$V = \pi r^2(\frac{S-2\pi r^2}{2\pi r})$

$\Rightarrow V = \frac{1}{2}(Sr - 2\pi r^3)$

Differentiating with respect r

$\frac{dV}{dr} = \frac{1}{2}(S-6\pi r^2)$

And $\frac{d^2V}{dr^2} = \frac{1}{2}(0-12\pi r)=-6\pi r$

For max and minima $\frac{dV}{dr}= 0$

$\frac{1}{2}(S-6\pi r^2)=0$

$\Rightarrow S = 6\pi r^2$ .

Since $\frac{d^2V}{dr^2} = -6\pi r<0$

Hence volume of cylinder is max when $S = 6\pi r^2$

Value of S putting in equation (i)

$h= \frac{6\pi r^2-2\pi r^2}{2\pi r}$

$\Rightarrow h = \frac{4\pi r^2}{2\pi r} = 2r$

Height of cylinder = Diametre of cylinder

Hence volume of cylindr is max when $h = 2r$

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