Minimize the material used in a cylinder

[Swallow]

Homework Statement

A cylinder with an open top is to be constructed with a predetermined volume of 1metre cube.
the base of the cylinder is to be cut from a sqaure sheet of material, and the rest of the square sheet is to be discarded, find the dimension which minimize the material used.

Homework Equations

Volume of cylinder = pi*R*R*H.

The Attempt at a Solution

The area of the base from which the square is to be cut is (2R)^2.
the area of the square wasted is ((2R)^2-(pi*R^2).
the surface area of the cylinder is 2*pi*R*H.
We have to minimize the function of (surface area of cylinder)+(Area of square wasted).
is this true?

 

[HallsofIvy]

No, you are asked to minimize "material used". By adding material needed for the curved side (what you call "surface area of the cylinder" is actually the area of the curved surface) and "area wasted", you have ignored the area used for the base. I see no reason to calculate the "area wasted" separately. If the cylinder has radius R and height H, then the material needed for the curved side is, as you say, 2\pi RH and material needed to make the base, both that used and wasted, is the area of the square, R^2. The total material needed is 2\pi RH+ R^2. Minimize that subject to the constraint that volume, \pi r^2H= 1.

 

[Liquidxlax]

you can use the lagrange multipliers, i think it works great if you've done partials


F = f(x,y) + \varphi = f(x,y) +\lambda(g(x,y)-c)


so the volume equals \varphi

and the surface area equals f(x,y) where x=r and y=h

take partials and use substitution and you'll get the right answer cause I did it on paper