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Question
Of all the closed cylindrical cans (right circular), of a given volume of 100 cubic centimetres, find the dimensions of the can which has the minimum surface area?
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Solution
Let r cm be the radius, h cm be the height, S cm2 be the total surface area and V cm3 be the volume.
Now,
V = πr2h = 100
⇒ `h = 100/(pir^2)`
and S = 2πr2 + 2πrh
⇒ `S = 2pir^2 + 2pir (100/(pir^2))`
`= 2pir^2 + 200/r`
Differentiate `S = 2pir^2 + 200/r` w.r.t r we get
`(dS)/(dr) = 4pir - 200/r^2`
For maximum / minimum surface area
`(dS)/(dr) = 0`
⇒ `4pir - 200/r^2 = 0`
⇒ `r^3 = 200/ (4pi)`
⇒ `r = (50/pi)^(1//3)`
`(d^2S)/(dr^2) = 4pi + (200 xx 2)/r^3`
`= 4pi + 400/r^3`
and `((d^2S)/(dr^2))_(r = (50/pi)^(1/3))`
`= 4pi + 400/ (50/pi) > 0`
∴ S has a minimum value at
`r = (50/pi)^(1/3)`
When `r = (50/pi)^(1/3)` cm, then
`h = 100/(pi(50/pi)^(2//3)) `
`h = 100/((50)^(2//3) pi^(1//3))`
`= (50xx2)/ ((50)^(2//3) pi ^(1//3))`
`= 2 (50/pi)^(1//3)` cm.
When r `(50/pi)^(1/3)` and h = 2 `(50/pi)^(1/3)` then S will be minimum.
Hence, the total surface area will be minimum.
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