Question

the slant height of a right cone is given as 10 cm. if the volume of cone is maximum, then the semi- vertical angle is:

• A. $-cos\frac{1}{\sqrt{3}}$
• B. $ta{n}^{-}1\sqrt{2}$
• C. $\frac{\pi }{6}$
• D. $\frac{\pi }{3}$

Answer 1

The correct option is A $-cos\frac{1}{\sqrt{3}}$

Volume of cone, $v=\frac{1}{3}\pi r^{2}h$

$\sin \alpha =\frac{r}{l}\Rightarrow r=l\sin \alpha$;

$\cos 2=\frac{h}{l}$

$\Rightarrow h=l\cos \alpha$

$\Rightarrow v=\frac{\pi }{3}\left(l^2{\sin }^2\alpha \right)\left(l\cos \alpha \right)=\frac{\pi l^3}{3}{\sin }^2\alpha \cos \alpha$

As $\frac{\pi l^3}{3}=$c(constant)

$\Rightarrow v=c(1-{\cos }^2\alpha )\cos \alpha$

$\Rightarrow v=c(\cos \alpha -{\cos }^3\alpha )$

For maxima, $\frac{dv}{d\alpha }=0$

$\Rightarrow -\sin \alpha +3{\cos }^2\alpha \sin \alpha =0$

$\Rightarrow \sin \alpha (3{\cos }^2\alpha -1)=0$

$\Rightarrow \sin \alpha =0$ (x)

${\cos }^2\alpha =\frac{1}{3}$

$\Rightarrow \cos \alpha =\frac{1}{\sqrt{3}}$

For $\cos \alpha >\frac{1}{\sqrt{3}},\frac{dv}{d\alpha }>0$

$\Rightarrow \alpha <\frac{1}{\sqrt{3}},\frac{dv}{d\alpha }>0$ & $\alpha >\frac{1}{\sqrt{3}},\frac{dv}{d\alpha }<0$

$\Rightarrow \alpha ={\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right)$ is maxima.