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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 $\large\frac{4r}{3}$

Toolbox:
• 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 $=r(1+\cos\theta)=r(1+\large\frac{1}{3})$
$\Rightarrow \large\frac{4r}{3}$
edited Sep 2, 2013