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Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius \(r\) is \( \large\frac{4r}{3}\)

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1 Answer

  • Area of a cone$=\large\frac{1}{3}$$\pi r^2h$
  • $\large\frac{d}{dx}$$\big(\sin \theta)=\cos\theta$
  • $\large\frac{d}{dx}$$\big(\cos \theta)=-\sin\theta$
Step 1:
The radius of sphere is $r$ in which a cone is inscribed .Let $O$ be the centre of the sphere.BC is the diameter of the base of cone.
Let $AM$=Altitude
$\angle BOM=\theta$
Radius of the cone $R=r\sin\theta$
Altitude of the cone ABC=$AM=AO+OM$
$\Rightarrow r+r\cos\theta=r(1+\cos\theta)$
We have Area of a cone$=\large\frac{1}{3}$$\pi r^2h$
By substituting the value of h we get
$\Rightarrow \large\frac{1}{3}$$\pi r^2h=\large\frac{1}{3}$$\pi r^2\sin^2\theta\times r(1+\cos\theta)$
$\Rightarrow \large\frac{1}{3}$$\pi r^2h=\large\frac{1}{3}$$\pi r^3\sin^2\theta\times (1+\cos\theta)$
Step 2:
Differentiating with respect to $\theta$
$\large\frac{dV}{d\theta}=\frac{1}{3}$$\pi r^3[2\sin\theta\cos\theta(1+\cos\theta)+\sin^2\theta(-\sin\theta)]$
$\quad\quad=\large\frac{1}{3}$$\pi r^3\sin\theta(2\cos\theta+2\cos^2\theta-\sin^2\theta)$
$\quad\quad=\large\frac{1}{3}$$\pi r^3\sin\theta(2\cos\theta+2\cos^2\theta-1+\cos^2\theta)$
$\quad\quad=\large\frac{1}{3}$$\pi r^3\sin\theta(3\cos^2\theta+2\cos\theta-1)$
$\quad\quad=\large\frac{1}{3}$$\pi r^3\sin\theta(\cos\theta+1)(3\cos\theta-1)$
Step 3:
$\large\frac{dV}{d\theta}$$=0$ at $\cos\theta=\large\frac{1}{3}$
$\cos\theta\neq -1$
Since $\theta\neq \pi$
$\large\frac{dV}{d\theta}$ changes sign from +ve to -ve.
$V$ is maximum at $\cos\theta=\large\frac{1}{3}$
Altitude $=r(1+\cos\theta)=r(1+\large\frac{1}{3})$
$\Rightarrow \large\frac{4r}{3}$
answered Aug 12, 2013 by sreemathi.v
edited Sep 2, 2013 by sharmaaparna1

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