Step 1:

Let $ABCDEF$ be a half cylinder with rectangular base and semi-circular ends.

Here $AB$=height of the cylinder

$AB=h$

Let $r$ be the radius of the cylinder.

Volume of the half cylinder is $V=\large\frac{1}{2}$$\pi r^2h$

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

$\therefore$ TSA of the half cylinder =LSA of the half cylinder+area of two semi-circular ends+area of the rectangle (base)=$\pi rh+\large\frac{\pi r^2}{2}+\large\frac{\pi r^2}{2}+$$h\times 2r$

$S=\pi r h+\pi r^2+2 h r$

$\;\;=(\pi r+2r)h+\pi r^2$

Substituting for h we get,

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

$\;\;=(\pi+2)\large\frac{2V}{\pi r^2}$$+\pi r^2$

Step 2:

Diff w.r.t $r$ we get,

$\large\frac{dS}{dr}=$$\bigg[(\pi+2)\times \large\frac{2V}{\pi}(\large\frac{-1}{r^2})+$$2\pi r\bigg]$

For maximum or minimum value of S,we have $\large\frac{dS}{dr}$$=0$

$\Rightarrow (\pi+2)\times \large\frac{2V}{\pi}(-\large\frac{1}{r^2})$$+2\pi r=0$

$\Rightarrow (\pi+2)\large\frac{2V}{\pi r^2}$$=2\pi r$

On simplifying we get,

$\large\frac{h}{2r}=\frac{\pi}{\pi+2}$

But $2r=D$

$\therefore h : D=\pi : \pi+2$

Step 3:

Diff $\large\frac{dS}{dr}$ w.r.t $r$ we get,

$\large\frac{d^2S}{dr^2}$$=(\pi+2)\large\frac{V}{\pi}\times \frac{2}{r^3}$$+2\pi > 0$

Thus S will be minimum when $h : 2r$ is $\pi : \pi-12$

(i.e) height of the cylinder : diameter of the circular end.

$\pi : \pi+2$