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# Show that the height of the cone of maximum volume that can be inscribed in a sphere of radius 12 cm is 16cm.

6 years ago
Think of a unit circle in polar coordinates, with the pole on the circle.

r = cos u; So what you want is to maximize

r^3(sin u)^2(cos u) that is (cos u)^4(sin u)^2.

If x = sin u, you want to maximize x(1-x^2) = x - x^3. The max is for x = 1/sqrt(3). Then (cos u)^2 = 1 -x^2 = 2/3.

So 2/3 * 24 = 16.
Arun Kumar IIT Delhi
6 years ago
Hi

$\\lets assume \theta is the angle made by joining \\center and one of the points on \\circumference of the cone base \\h_{cone}=r+rcos\theta \\r_{cone}=rsin\theta \\v_{cone}={\pi r_{cone}^2h_{cone} \over 3} \\=>lnv_{cone}=ln(constant)+2lnr_{cone}+lnh_{cone} \\=>0=0+2/r_{cone}+{dh_{cone} \over h_{cone}dr } \\=>{dh_{cone} \over dr_{cone}}=-2{h_{cone} \over r_{cone}}=-2{r+rcos\theta \over rsin\theta}={-rsin\theta \over rcos\theta} \\=>2(1+cos\theta)cos\theta=sin^2\theta \\=>2cos\theta+2cos^2\theta=1-cos^2\theta \\=>3cos^2\theta+ 2cos\theta-1=0 \\=>cos\theta={-2 \pm \sqrt16 \over 6}=1/3(possible \,value) \\=>h=4r/3=16$

Thanks & Regards, Arun Kumar, Btech,IIT Delhi, Askiitians Faculty