Question
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.
Answers
Steve
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
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