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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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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.

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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.
answered Aug 9, 2013 by sreemathi.v
 

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