Find the dimensions of a cylindrical can that will minimize the cost of the material

Question

A cylindrical can is to be made to hold 1 L of oil. Find the dimensions that will minimize the cost of the metal to manufacture the can.

This is a standard calculus problem.

The volume of the cylinder is:

V == π r^2 h

The area of the cylinder is:

A[r_] := 2 π r^2 + 2 π r h /. h -> V/(π r^2)

Find r when the slope of the area is zero:

Reduce[A'[r] == 0]

Result:

(r == -(-(1/(2 π)))^(1/3) V^(1/3) || 
   r == V^(1/3)/(2 π)^(1/3) || 
   r == ((-1)^(2/3) V^(1/3))/(2 π)^(1/3)) && r != 0

As you can see, when I defined A[r], I replaced h with an expression produced by solving the volume equation for h.

My question is, is there a way to express the problem in terms of Reduce and a set of equations, without the manual solving for h?

I.e. something along the lines of:

A[r_] := 2 π r^2 + 2 π r h

Reduce[{A'[r] == 0, V == π r^2 h}]

Of course, that doesn't yield the correct answer because A'[r] doesn't treat h as being in terms of r.

Answer 1

As requested, here is a method that uses Reduce[]:

Reduce[Dt[2 Pi r^2 + 2 Pi r h, r] == 0 && Pi r^2 h == 1/1000 && Positive[r], {r, h}]

I'd still prefer using Minimize[], though.

Answer 2

OK, I found one approach:

A[r_] := 2 \[Pi] r^2 + 2 \[Pi] r h
V[r_] := \[Pi] r^2 h
Reduce[{Dt[A[r], r] == 0, V[r] == 1, r > 0}]

Result:

r == 1/(2 \[Pi])^(1/3) && h == 2^(2/3)/\[Pi]^(1/3)