From all the closed right circular closed cylindrical can of volume ` 128 pi cm^3`, Find the dimension of can which has minimum surface area.

Explanation

Here, we are given the volume of cylindrical can `128picm^3`.

`:. pir^2h = 128pi => h = 128/r^2`

Now, surface area of cylinder, `S = 2pir^2+2pirh`

`=> S= 2pir^2+2pir(128/r^2)`

`=> S= 2pir^2+ (256pi)/r`

For minimum value of surface area, `(dS)/(dr)` should be `0`.

`=>(dS)/(dr) = 4pir+256pi(-1/r^2) = 0`

`=> 4pir = (256pi)/r^2=>r^3 = 64=> r = 4cm`

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