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Show that the cylinder of given volume open at the top has minimum total surface area if its height is equal to radius of the base.

1 Answer

Toolbox:
  • $V=\pi r^2h$
Step 1:
Let $r$ be the radius and $h$ be the the height of a cylinder of given volume $V$
$V=\pi r^2h$
$h=\large\frac{v}{\pi r^2}$
Let $S$ be total surface area.Since the cylinder is open at its top.
$S=2\pi rh+\pi r^2$
Substituting for $h$ we get,
$S=2\pi r\big(\large\frac{v}{\pi r^2}\big)$$+\pi r^2$
$\;\;=\large\frac{2v}{r}$$+\pi r^2$
Step 2:
Differentiating with respect to $x$ we get,
$\large\frac{dS}{dr}=\frac{-2v}{r^2}$$+2\pi r$
For maximum or minimum
$\large\frac{dS}{dr}=$$0$
$\Rightarrow \large\frac{2v}{r^2}+$$2\pi r=0$
(i.e) $\large\frac{-v}{r^2}$$+\pi r=0$
$\therefore v=\pi r^3$
$\pi r^2h=\pi r^3$
$\Rightarrow h=r$
Step 3:
Differentiating again with respect to $r$ we get,
$\large\frac{d^2S}{dr^2}=\frac{4v}{r^3}$$+2\pi$
When $r=h$
$\large\frac{d^2S}{dr^2}=\frac{4v}{r^3}$$+2\pi$ >0
Hence $S$ is minimum when $h=r$ (i.e) when the height of the cylinder is equal to the radius of the base.
answered Oct 2, 2013 by sreemathi.v
 

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