Browse Questions

# Show that the right-circular cone of least curved surface and given volume has an altitude equal to $\sqrt 2$ times the radius of the base.

Toolbox:
• $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) $k^2=h^2r^4$
$h^2r^4=2r^6$
$h^2=2r^2$
$h=\sqrt{2r^2}$
$h=r\sqrt 2$