




Minimize the surface area A of a tin can of fixed volume V.
Area of the tin can: A = 2pr^{2 }+ 2prh
Volume of the tin can: V = pr^{2}h
We introduce a scaleinvariant variable (dimensionless quantity):
We express h and r as function of V and t:
We express A as a function of V and t:
V is fixed and therefore min A is attained for C(t) minimum, that is for t^{2/3} + t^{1/3} minimum.
Plotting of:
y = t^{2/3} + t^{1/3}
Minimum of f(t) = t^{2/3} + t^{1/3}
The minimum of f(t) is obtained for df(t) / dt = 0
The surface area A of a tin can of fixed volume is minimized when h = 2r.