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# The sum of the surface areas of  a sphere and a cube is given.Show that when the sum of their volumes is least, the diameter of the sphere is equal to the edge of the cube.

Arun Kumar IIT Delhi
7 years ago
Hi

$\\4\pi r^2+6a^2=p \\ {4\pi r^3 \over 3}+a^3=v \\=>4\pi r^2+6a^2=p \\=>8\pi 2r{dr \over dt }+12a{da \over dt}=0 \\=>{dr \over da}={-12a \over 16\pi r} \\v ={4\pi r^3 \over 3}+a^3 \\=>{dv \over da}=4\pi r^2{dr \over da}+3a^2=0$

$\\=>4\pi r^2*{-12a \over 16\pi r}+3a^2=0 \\=>-3ar+3a^2=0 \\=>r=a$
there could be a mistake in the question it should be actually radius.

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