Question

If $θ$ is the semi-vertical angle of a cone of maximum volume and given slant height, then $\tan θ$ is given by

• A. $2$
• B. $1$
• C. $\sqrt{2}$
• D. $\sqrt{3}$

Answer 1

Answer: (C)

Volume of cone, $V=\frac{Ï€}{3}{r}^{2}h$
$⇒V=\frac{π}{3}{r}^2\sqrt{l^2−r^2}$

On differentiating w.r.t. r we get
$\frac{dV}{dr}=\frac{π}{3}\left[2r\sqrt{l^2−r^2}+\frac{r^2}{2\sqrt{1^2−r^2}}(−2r)\right]$
Put $\frac{dV}{dr}=0$, we get
$r=Â\pm l\sqrt{\frac{2}{3}}$
At $r=l\sqrt{\frac{2}{3}},\frac{d^{2}V}{d{r}^2}<0,$ (maximum)
$∴h=\sqrt{l^2−\frac{2}{3}{l}^2}−\frac{l}{\sqrt{3}}$
In $ΔABC,\tan θ=\frac{r}{h}=\frac{l\sqrt{\frac{2}{3}}}{l/3}=\sqrt{2}$