Q. Show that a closed right circular cylinder of given total surface area S and maximum volume V is such that its height h is equal to diameter of the base.

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S is the total surface area of the closed right circular cylinder whose radius is r, height is h, and volume is V.

Then surface area `S= 2pir^2+2pirh`

`h= (S-2pir^2)/(2pir)` ------ (i)

Volume `V = pir^2h=pi[(S-2pir^2)/(2pir)]= 1/2[Sr-2pir^3]`

Therefore, `(dV)/(dr)=1/2[S-6pir^2]`

For extrema (maxima or minima) dV/dr=0

Therefore `S-6pir^2=0`

`rArr S = 6pir^2`

Plugging this value of S in eq.(i),we get

h (at V-extrema)=`(6pir^2-2pir^2)/(2pir) =2r `

For maxima, `(d^2V)/(dr^2)= -6pir` = negative

Therefore, V is maximum when height, **h = 2r** = diameter of the base.

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