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A given quantity of metal is to be cast into a solid half cicular cylinder (i.e., with rectangular base and semicircular ends). Show that in order that the total surface area may be minimum, the ratio of the length of the cylinder to the diameter of its circular ends is \( \pi : ( \pi + 2). \)

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  • Volume of the half cylinder is $V=\large\frac{1}{2}$$\pi r^2h$
Step 1:
Let $ABCDEF$ be a half cylinder with rectangular base and semi-circular ends.
Here $AB$=height of the cylinder
Let $r$ be the radius of the cylinder.
Volume of the half cylinder is $V=\large\frac{1}{2}$$\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}$$=2\pi r$
On simplifying we get,
But $2r=D$
$\therefore h : D=\pi : \pi+2$
Step 3:
Diff $\large\frac{dS}{dr}$ w.r.t $r$ we get,
$\large\frac{d^2S}{dr^2}$$=(\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.
$\pi : \pi+2$
answered Nov 8, 2013 by sreemathi.v

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