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A cylindrical can is made from tin.If it can be contain liquid inside it then what is the parameter of design if we are oblige use the minimum amount of tin.

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assuming a constant volume v, we have

v = pi r^2 h
so, h = v/(pi r^2)

the surface area is

a = 2pi r(r+h)
= 2pi r(r + v/(pi r^2))
= 2 pi r^2 + 2v/r

da/dr = 4pi r - 2v/r^2
= 2(2pi r^3 - v)/r^2

we want da/dr=0 for max/min area, so

r = ∛(v/(2pi))
h = v/(pi r^2) = ∛(4v/pi)

but, is this min or max area? Check a'' to be sure it's a minimum

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