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Question
Show that the right circular cylinder of given surface and maximum volume is such that is heights is equal to the diameter of the base.
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Solution
Let r be the radius of the circular base, h be the height and S be the total surface area of a right circular cylinder, Then S = 2πr2 + 2πrh.
Let V be the volume of the cylinder with the above dimensions.
∴ `V = pir^2h = pir^2 ((S - 2pir^2)/(2pir))`
`(∵ S = 2pir^2 + 2pirh, ∴ h = (S - 2pir^2)/(2pir))`
`= r/2 (S - 2pir^2)`
⇒ `V = (sr)/2 - pir^3`
Differentiating w.r.t. x, we get
`(dV)/(dr) = S/2- 3pir^2`
For maximum / minimum volume
`(dV)/(dr) = 0`
⇒ `S/2-3pir^2 = 0`
⇒ `r^2 = S/(6pi)`
⇒ `r = sqrt(S/(6pi))`
`(d^2V)/(dr^2) = -6pir`
and `((d^2V)/(dr^2))_(r sqrt (S/(6pi)))`
`= -6pi sqrt (S/(6pi)) < 0`
⇒ V has a maximum value at `r = sqrt (S/ (6pi))`
When `r = sqrt (S/ (6pi)), `then
`h = (S- 2pi (S/(6pi)))/ (2pi sqrt (S/ (6pi))) = (4pi (S/ (6pi)))/ (2pi sqrt (S/ (6pi)))`
⇒ `h = 2 sqrt (S/(6pi)) = 2` radius = diameter.
So volume is maximum when the height is equal to the diameter of the base.
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