Question

Show that the height of a cylinder, which is open at the top, having a given surface area and greatest volume, is equal to the radius of its base. 

Answer 1

Given a cylinder 

S = 2πrh + πr²             .......(1)

V = πr² h                    .......(2 )

From (1) S - πr² = 2 πrh

`h = [(S -pir^2)/(2pir)]  = S/(2pir) - r/2`

Put in (2) 

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

Now diff on both sides by ‘r’

`(dv)/(dr) = S/2 - (3pir^2)/2`

For max/min `(dv)/(dr) = 0`

`S/2 - (3pir^2)/2 = 0 ⇒ S = 3pir^2`      ........(3)

`(d^2 v)/(dr^2) = -3pir = -3 pi x sqrt(S/3pi) <0`

∴ By second derivative test it is maxima from (1) & (3) 

`2pirh + pir^2 = 3pir^2`

`2pir h = 2 pir^2`

h = r