The sum of the surface areas of a cuboid with sides `x ,2x` and `x/3` and a sphere is given to be constant. Prove that the sum of their volumes is minimum, if `x` is equal to three times the radius of sphere. Also find the minimum value of the sum of their volumes.

Explanation

Given `quad S=4 pi r^{2}+2[frac{x^{2}}{3}+2 x^{2}+frac{2 x^{2}}{3}]`



`S=4 pi r^{2}+6 x^{2} {Or} x^{2}=frac{S-4 pi r^{2}}{6} `


`text { and } quad V =frac{4}{3} pi r^{3}+frac{2 x^{3}}{3} `


`therefore quad V=frac{4}{3} pi r^{3}+frac{2}{3}(frac{S-4 pi r^{2}}{6})^{3 / 2} `


`frac{d V}{d r} =4 pi r^{2}+(frac{S-4 pi r^{2}}{6})^{1 / 2}(frac{-8 pi r}{6}) `


`frac{d V}{d r}= 0`



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