# Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is$$\large\frac{8}{27}$$ of the volume of the sphere.

Toolbox:
• Volume of cone $=\large\frac{1}{3}$$\pi r^2h • \large\frac{d}{d\theta}$$\sin\theta=\cos\theta$
Step 1:
Let a cone $VAB$ of greatest volume be inscribed in the sphere.
Let $AOC=\theta$
$AC$,radius of the base of the cone $=R\sin\theta$
$VC=VO+OC$
$\;\;\;\;\;\;=R+R\cos\theta$
$\;\;\;\;\;\;=$height of the cone
$V$-The volume of the cone.
$\Rightarrow \large\frac{1}{3}$$\pi (AC)^2VC Substitute the value of AC and VC \Rightarrow V=\large\frac{1}{3}$$\pi R^2\sin^2\theta(R+R\cos\theta)$
$\Rightarrow V=\large\frac{1}{3}$$\pi R^2\sin^2\theta R(1+\cos\theta) \Rightarrow V=\large\frac{1}{3}$$\pi R^3\sin^2\theta (1+\cos\theta)$
Step 2:
Differentiating with respect to $\theta$ we get,
$\large\frac{dV}{d\theta}=\large\frac{1}{3}$$\pi R^3[\sin^2\theta(-\sin \theta)+(1+\cos\theta).2\sin\theta\cos\theta] \qquad=\large\frac{1}{3}$$\pi R^3[-\sin^3\theta+2\sin\theta(1-\sin^2\theta)+2\sin\theta\cos\theta]$
$\qquad=\large\frac{1}{3}$$\pi R^3[-3\sin^3\theta+2\sin\theta+2\sin\theta\cos\theta] For maximum and minimum we have \large\frac{dV}{d\theta}$$=0$
$\Rightarrow \large\frac{1}{3}$$\pi R^3[-3\sin^3\theta+2\sin\theta+2\sin\theta\cos\theta]=0 \Rightarrow -3\sin\theta+2+2\cos\theta=0[\sin\theta\neq 0] \Rightarrow -3(1-\cos^2\theta)+2+2\cos\theta=0 \Rightarrow 3\cos^2\theta+2\cos\theta-1=0 (3\cos\theta-1)(\cos\theta+1)=0 \cos\theta=\large\frac{1}{3} \cos\theta=-1 Step 3: But \cos\theta=-1 as \cos\theta=-1 \Rightarrow \theta=\pi which is not possible \cos\theta=\large\frac{1}{3} When \cos\theta=\large\frac{1}{3} \sin\theta=\sqrt{1-\cos^2\theta} \qquad\;=\sqrt{1-1/9}=\sqrt{\large\frac{9-1}{9}} \qquad\;=\large\frac{2\sqrt 2}{3} \large\frac{dV}{d\theta}=\frac{1}{3}$$\pi r^3\sin\theta[-3\sin^2\theta+2+2\cos\theta]$
$\qquad=\large\frac{1}{3}$$\pi r^3\sin\theta[3\cos\theta-1](\cos\theta+1)=0 Step 4: When \theta is slightly < \cos^{-1}\big(\large\frac{1}{3}\big) \sin\theta =+ve 3\cos\theta-1=+ve[as \theta decreases,\cos\theta increases] \cos\theta+1=+ve \large\frac{dV}{d\theta}$$=(+)(+)(+)\Rightarrow +ve$
When $\theta$ is slightly > $\cos^{-1}\big(\large\frac{1}{3}\big)$
$\sin\theta =+ve$
$\cos\theta+1=-ve$
[Since as $\theta$ increases the value of $\cos\theta$ decreases]
$\large\frac{dV}{d\theta}$$=(+)(-)(+)\Rightarrow -ve \large\frac{dV}{d\theta} changes from +ve to -ve. Step 5: Hence V is maximum at \theta=\cos^{-1}\big(\large\frac{1}{3}\big) Maximum volume of cone =\large\frac{1}{3}$$\pi R^3\sin^2\theta(1+\cos\theta)$
Now $\cos\theta=\large\frac{1}{3}$
$\sin\theta=\large\frac{2\sqrt 2}{3}$
Step 6:
Therefore Maximum volume of cone=$\large\frac{1}{3}$$\pi R^3\big(\large\frac{2\sqrt 2}{3}\big)^2$$\big(1+1/3\big)$
$\Rightarrow \large\frac{1}{3}$$\pi R^3\large\frac{8}{9}\times \frac{4}{3} \Rightarrow \large\frac{32}{81}$$\pi R^3$
$\Rightarrow \large\frac{8}{27}\big(\frac{4}{3}$$\pi R^3\big)$
$\Rightarrow \large\frac{8}{27}\times$ volume of the sphere.