Max Volume of Cube from 1000 cm^2 Sheet

[physicsnewb7]

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?

 

[berkeman]

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?

Is the answer something other than the obvious one?

 

[DaveC426913]

I am going to specify that the sheet is 1mm wide and 100m long.

This allows me to construct the cube from ribbons without wasting any material, meaning my cube will be five panels covering an area of 1000cm^2, so 200cm^2 each, so 14.14cm on a side - for a volume of 2828.43cm^3.

Anyone beat me?

 

[uart]

Anyone beat me?

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 :

$$v = 250 x - 0.25 x^3$$

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.

 

[DaveC426913]

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 :

$$v = 250 x - 0.25 x^3$$

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.

While this is great and everything:
- Is it a cube? It looks to me like it's a prism 18cm on a side and only 9cm deep.
- what does your layout look like on the 1000cm^2 sheet?
- Can you actually build it from the sheet?

My answer should be the largest possible volume (i.e. it should at least equal, if not better, yours). If it isn't, then it's me who's made the mistake in my calcs. See, the only thing I needed to solve was how to maximize the amount of material used from the original sheet i.e. no wastage.

So, let me check my numbers: What is the largest possible cube whose 5 of 6 sides total 1000cm^2? Shouldn't it be only 200cm^2 per face? Which means each face can only be sqrt(200)cm on a side (14.14). How how you get bigger faces?

 

[DaveC426913]

I have a better answer:

I can make a cube with a volume of 1,000,000,000cm^3, or 1,000m^3.

The other thing not specified in the OP is the thickness of the material. I have chosen a material that is 1000cm thick, meaning I can make a cube from it that is 1000cm on a side.

 

[uart]

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).

 

[DaveC426913]

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).

He specified cube twice.

You stated your assumptions in your solution, granted, but you've redefined "cube":

"box/cube" means "box" (rectangular prism) or "cube"

i.e. box/cube = box = rectangular prism = cube. You have ignored the fact that the rectangular prism's height must equal its length/width to be able to call it a cube.

 

[uart]

Yes that's true I cheated in order to make it a mathematically more interesting problem.

Though I'm honestly still tending to think that the OP was erroneously using the term "cube" when he actually meant "rectangular prism" (yes I understand that they're not the same thing).

 

[physicsnewb7]

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

 

[uart]

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 : $\sqrt{(200)}$ 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?

 

[DaveC426913]

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 : $\sqrt{(200)}$ 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?

Yes, I think there's still some ambiguity as to constraints. For example, The xy dimensions of the material are unspecified, so my first answer assumes a extremely thin ribbon, allowing me complete freedom and no waste. Secondly, there's the questions of how one may cut up the material to form the sides.