Show that the total surface area of a closed cuboid with square base and given volume is minimum, when it is a cube.

Explanation

The sides of the cuboid in the square base can be `x, x` and `y`


Let total surface area `=S=2 x^{2}+4 x y`


As volume of cuboid is `V=x^{2} y=` constant


`therefore quad y=frac{V}{x^{2}}`


`therefore quad S=2 x^{2}+4 x cdot frac{V}{x^{2}}=2 x^{2}+frac{4 V}{x}` [Substituting `(` ii `)` in `.(i)]`


To find condition for minimum `S` we find `frac{d S}{d x}`


`Rightarrow quad frac{d S}{d x}=4 x-frac{4 V}{x^{2}}`


If `frac{d S}{d x}=0 Rightarrow quad 4 x^{3}=4 {V}`


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