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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Show that the right circular cone of least curved surface and given volume has an altitude equal to \(\sqrt 2\) time the radius of the base.

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  • $V=\large\frac{1}{3}$$\pi r^2h$
Step 1:
Let $r$ and $h$ be the radius and height of the cone respectively.
Volume $V=\large\frac{1}{3}$$\pi r^2h$
$\qquad=\large\frac{\pi k}{3}$$(constant)$
$r^2h=k$ or $h=\large\frac{k}{r^2}$------(1)
Surface $S=\pi rl=\pi r(\sqrt{h^2+r^2})$
$h=\large\frac{k}{r^2}$from (1)
$S=\pi r\sqrt{\large\frac{k^2}{r^4}+r^2}$
$\quad=\pi r\large\frac{\sqrt{k^2+r^6}}{r^4}$
$\quad=\pi \large\frac{\sqrt{k^2+r^6}}{r}$
Step 2:
$\large\frac{dS}{dr}$$=\pi\bigg[\large\frac{\Large\frac{6r^5}{2\sqrt{r^6+k^2}}\times r-\sqrt{r^6+k}.1}{r^2}\bigg]$
$\quad=\large\frac{3r^6-(r^6+k^2)}{r^2(\sqrt{r^6+k^2})}$
$\quad=\large\frac{(2r^6+k^2)}{r^2(\sqrt{r^6+k^2})}$
$\quad=k^2=2r^6$
$\large\frac{dS}{dr}$$=0$
Step 3:
$\large\frac{dS}{dr}$ changes sign from -ve to +ve as $r$ increases through the point $k^2=2r^6$
$\Rightarrow S$ is the least at this point.
From (1) $k^2=h^2r^4$
$h^2r^4=2r^6$
$h^2=2r^2$
$h=\sqrt{2r^2}$
$h=r\sqrt 2$
answered Aug 9, 2013 by sreemathi.v
edited Aug 30, 2013 by sharmaaparna1
 

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