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Question

Show that the height of a closed right circular cylinder of given surface and maximum volume, is such that its height is equal to the diameter of its base.

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Solution

Let S be the given surface area of the closed cylinder where radius is r and height is h.
Let V be the volume
Surface area S=2πr2+2πrh
h=S2πr22πr ....(1)
Volume =πr2h
Substitute the value of h in above equation,
Volume =πr2[S2πr22πr]
=12r(S2πr2)
=12(Sr2πr3)
Differentiating wrt r, we get
dVdr=12(S6πr2)
For maxima and minima dVdx=0
S6πr2=0
S=6πr2
From (1), we havee
h=S2πr22πr
Substitute the value of S in the above equation,
h=6πr22πr22πr
=4πr22πr
Thus h=2r
On double differentiation of V, we get
d2Vdr2=12(12πr)
=6πr
ve
Therefore, V is maximum.
Thus volume is maximum when h=2r.
Or when height of cylinder is equal to diameter of base.

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