Prove that the volume of the largest cone that can be

21/02/2023 by Team Gmath

Question: Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is 8/27 of the volume of the sphere.

Solution: Consider $O$ be the centre and $R$ be the radius of the given sphere, $O M=x$

In ΔOBM, using pythagoras theorem

$OB^2=OM^2+BM^2$

$\Rightarrow R^2 = x^2 + BM^2$

$\Rightarrow BM^2 = R^2-x^2$

$AM= h = R+x$

BM is the radius of cone

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

$\Rightarrow V = \frac{1}{3}\pi(R^2-x^2)(R+x)--(i)$

$\Rightarrow V = \frac{1}{3}\pi(R-x)(R+x)(R+x)$

$\Rightarrow V = \frac{1}{3}\pi (R-x)(R+x)^2$

Differentiating with respect to x

$\frac{dV}{dx} = \frac{1}{3}\pi [(R-x)2(R+x)+(R+x)^2(-1)]$

$= \frac{1}{3}\pi (R+x)[2(R-x)-(R+x)]$

$= \frac{1}{3}\pi(R+x)[2R-2x-R-x]$

$=\frac{1}{3}\pi (R+x)[R-3x]$ —(ii)

For max and minima $\frac{dV}{dx}=0$

$\frac{1}{3}\pi (R+x)[R-3x]=0$

Hence $R+x=0$ or $R-3x = 0$

$x = -R$ or $x = R/3$

$x =-R$ Not possible

Agian differentiate with respect to x of eq (ii)

$\frac{d^2V}{dx^2} = \frac{1}{3}\pi[(R+x)(-2)+(R-2x).1]$

$= \frac{1}{3}\pi[-2R-2x+R-2x]$

$= \frac{1}{3}\pi[-R-4x]$

At $x = R/3$

$\frac{d^V}{dx^2}=\frac{1}{3}\pi[-R-4\times \frac{R}{3}]$

$= \frac{1}{3}[-7R/3]<0$

Hence volume of cone is max when $x=R/3$

Now From eq (i)

$V = \frac{1}{3}\pi(R^2-\frac{R^2}{9})(R+\frac{R}{3})$

$= \frac{1}{3}\pi \frac{8R^2}{9}.\frac{4R}{3}$

$= \frac{8}{27}.\frac{4}{3}\pi R^3$

Volume of cone $= \frac{8}{27}\times$ volume of sphere

https://gmath.in/a-wire-of-length-28-m-is-to-be-cut-into-two-pieces/

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