logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XII  >>  Math  >>  Model Papers
0 votes

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 \( \pi : ( \pi + 2). \)

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • Volume of the half cylinder is $V=\large\frac{1}{2}$$\pi r^2h$
Step 1:
Let $ABCDEF$ be a half cylinder with rectangular base and semi-circular ends.
Here $AB$=height of the cylinder
$AB=h$
Let $r$ be the radius of the cylinder.
Volume of the half cylinder is $V=\large\frac{1}{2}$$\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.
$\pi : \pi+2$
answered Nov 8, 2013 by sreemathi.v
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...