Symmetrical Proof

[AUuserID]

Can someone help me with a proof for symmetry in trying to maximize the volume of a box, or a general form would be even better. The box is in its most simple form V=xyz, with a limited amount of surface area. First the eqeuation must be solved in one dimension to reduce it to two variables then partial derivation may be used to solve the equation. My assumption is that the volume would be maximized, (IF you have a limited surface area to work with) if you let the sides be square; i.e. x=y. I know this may seem obvious but is there any general mathematicalproof for this?

[jemidiah]

I remember seeing this (or it may have been a square with a limited length and width, although they would be almost the same) from geometry class. I do remember that the answer is L=H=W, but I, unfortunately, don't remember the proof. The only thing I can suggest is that you could try searching around the internet (http://www.google.com/ ;)).

[Guv]

The 2D version of this is easy to prove.

For a given perimeter of a rectangle, the following is true. (X + Y) = Perimeter/2, if (X, Y) are the lengths of sides. For a square, Side = Perimeter/4 and Area = Side2 For rectangle, X = (Side - Delta) and Y = (Side + Delta), where Side is the length of a side of the square. This is required, since X + Y must equal (Side + Side). XY = (Side - Delta)(Side + Delta) Hence XY = Side2 - Delta2 < Side2I do not see an obvious proof for a rectangular parallelopiped, but my intuition tells me that a cube would have the maximum volume for a given surface area.

Using MathCad7, I did some numerical experiments which reinforced my intuition.

[Guv]

BTW: There is a known proof that a sphere has maximum volume for a given surface area.

If the surface area is known, Radius2 = Area/4Pi defines the radius of a sphere with that surface area. No other 3D shape with that surface area has more volume than 4*Pi*Radius3/3

Similarly, there is a proof for 2D shapes that the circle has maximum area for a given perimeter.

I wonder if the circle proof might provide a hint for the Cube/Rectangle problem.