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A sheet of material 1000 cm^2 is given to you to construct a box/cube with no lid. What is the maximum volume of the cube that can be created in cm^3?
A sheet of material 1000 cm^2 is given to you to construct a box/cube with no lid. What is the maximum volume of the cube that can be created in cm^3?
Anyone beat me?
While this is great and everything:Sure can.
I'm going to assume that "box/cube" means "box" (rectangular prism) or "cube" so I'll maximize v = x y z subject to xy + 2xz + 2yz = 1000.
This can be greatly simplified by noting the symmetry with respect to x and y (the dimensions of the base) and hence assume that the optimum solution will have x=y.
Maximize v=x^2 z subject to x^2 + 4xz = 1000.
Rearranging the constraint and substituting into "v" gives :
[tex] v = 250 x - 0.25 x^3[/tex]
Maximizing in the usual manner gives x = y = 18.257 cm and z=x/2 = 9.129cm (both approx to 3dp) .
BTW. This results in an aprox volume of V = 3042.9 cm^3.
He specified cube twice.Yes as per usual (for most new users here) it's an incredibly poorly specified problem. The shape of the "sheet" of material was completely unspecified so just make it whatever shape you require for the net of your solution.
As for the box/cube part I genuinely believe that the he meant "box" and the cube part was optional. That is, I think that this was actually meant to be just your typical volume/SA maximization problem (as per my solution).
Yes, being a new user here I lacked certain specifications that would indeed clarify the problem. When I said "material" I meant it to be infinitely thin, like a plane. And the material is perfectly malleable. And when I refer to "box/cube", I mean it must have all equal faces or sides. I didn't want to use just the word cube because it isn't really a cube since there is no top. Thats why I stated a hybrid between box and cube. I apologize for the confusion and appreciate the responses
That's ok physicsnewb7.
It just means that mathematically it's not really a very interesting problem, if it must be a cube, since the solution is a little too obvious : [itex]\sqrt{(200)}[/itex] cm per side.
Or was the problem to find a solution whos net could be cut neatly from a rectangular sheet of material (that is,without introducing any extra seams). Can you please elaborate on this?