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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.
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Solution
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
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Perimeter of rectangle `=2(x+y)=2(x+50/x)`
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When x = `root(5)(2),y=50/root(5)(2)=root(5)(2)`
∴ `x=root(5)(2) "cm" , y = root(5)(2) "cm"`
Hence, rectangle is a square of side `root(5)(2) "cm"`
