Minimize the surface area A of a tin can of fixed volume V.
Area of the tin can: A = 2pr2 + 2prh
Volume of the tin can: V = pr2h
We introduce a scale-invariant 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 + t1/3 minimum.
y = t-2/3 + t1/3
Minimum of f(t) = t-2/3 + t1/3
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.