1.

Showthat the height of a closed right circular cylinder of given surface andmaximum volume, is equal to the diameter of its base.

Answer» Surface area of cylinder ` A = 2pir(r+h)`
`A = 2pir^2+2pirh`
`h = (A-2pir^2)/(2pir)`
Now, volume of cylinder `V = pir^2h`
`=>V = pir^2((A-2pir^2)/(2pir))`
`=>V = (Ar-2pir^3)/2`
For maximum volume,
`(dV)/(dr) = 0`
`:. A/2-3pir^2 = 0`
`=>3pir^2 = A/2`
`=>6pir^2 = A->(1)`
Now, As `A = 2pir^2+2pirh`
`:. 2pir^2+2pirh = 6pir^2`
`=>2pirh = 4pir^2`
`=>h = 2r`
So, height of given cylinder is double of the radius.
In other words, height is equal to the diameter of the cylinder.


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