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

Explanation

Volume of a cuboid witha square base `V= l**l**h`

Here, `l` is length and breadth and `h` is the height.

`:. V = l^2h => h = V/l^2`

Now, total surface area of the cuboid `S = 2(l**l+l**h+l**h) `

`=> S = 2l^2+4lh`

`=> S = 2l^2+4l(V/l^2)`

`=> S = 2l^2+(4V)/l`

Now, surface area will be minimum when `(dS)/(dl)` is `0`.

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