STD 12 - 6. Application of derivatives — JEE STD 11 Science — Question

Let height and radius of cone is $h$ and $r$ respectively, $h, r \in I$

Given volume of cone $=$ Surface area of cone

$\frac{1}{3} \pi r^2 h=\pi r l+\pi r^2$

$\frac{1}{3} \pi r^2 h=\pi r \sqrt{h^2+r^2}+\pi r^2$

$\frac{1}{3} r h=\sqrt{h^2+r^2}+r \quad[r \neq 0]$

$r h-3 r=3 \sqrt{h^2+r^2}$

$\Rightarrow r^2 h^2+9 r^2-6 h r^2=9 h^2+9 r^2$

$\Rightarrow h^2\left(r^2-9\right)=6 h r^2$

$h =\frac{6 r^2}{r^2-9}$

$h =6\left(\frac{r^2}{r^2-9}\right)$

$h =6+\frac{54}{r^2-9}$

$h$ and $r$ are integer.

$\because r^2-9$ is a factor of 54 .

$\therefore r^2-9=1,2,3,6,9,18,27,54$

$r^2 =10,11,12,15,18,27,36,63$

$r =6 \text { only possible value }$

$h =6+\frac{54}{36-9}$

$=6+2=8$

$r =6, h=8$