How many minimally sides are needed to fully enclose a volume in an $n$-dimensional spaece? Imagine I have some unbendable*, but cuttable metal-plates. One side is silver, the other one is blue. 
Now I want to use them to build somekind of case (with no opening) that is fully blue. It just needs to keep the air inside without any chance of it going out for some reason, and it needs to be blue. You cannot use the silver-side to contact anything but the inside of the structure.
Also, the metal used is very, very expensive, so I need to use as little as possible (but the structure's volume $v$ is given).
How many pieces of metal do I need to enclose a certain volume of air in an n-dimensional space?
I, a layman in math, discussed this with my friend who is quite good at math, and he pointed my attention to one thing:
If you have an one-dimensional space, you'd need two points to enclose everything in-between. If you'd have a two-dimensional space, you'd need 3 to enclose anything in-between (building a triangle). And in three-dimensional space, we both believe the tetrahedron is (is it really? can this be proven?) the best structure for this, so we'd need 4 metal plates.
But my company expands further into the universe right now, and four-dimensional aliens need our product too. It's my task to calculcate how many metal places are minimally needed. And since we want to expand even further, I'd like to have fifth, sixth, seventh, ..., $n$th dimensional spaces too.
Is there any way to proof that it's always $n + 1$ plates in $n$ dimensions? Or is there any better way to do that?
Thanks.
*It's not the year 3000, so no, Bender Rodriguez cannot bend them for us.
 A: Yes, the minimal number of linear subspaces of dimension $n-1$ that encloses a non-zero, finite
$n$-volume in $n$-space is always $n+1$.  
However, you have missed a point, which I will illustrate in 2 dimensions:  For a fixed volume of $1$, the triangle with least perimeter that encloses a volume of $1$ has perimeter 
$$
\frac{6}{\sqrt[4]{3}}\approx 4.56$$
But a better choice if the metal is precious is a square, with perimeter $4$.  A still better choice is a regular hexagon, with perimeter 
$$2\sqrt{2}\sqrt[4]{3} \approx 3.72$$
The point is that for any perimeter limit greater than $2\sqrt{\pi}$ you can find a regular polygon with a perimiter less than that limit, enclosing a volume  (area) of $1$.  So the problem as posed has no solution (if Bender is not permitted).  The same situation holds for all dimension greater than $1$.
A much more interesting problem arises if you say that the metal is precious but so are the hinges joining two sides.  For example, in two dimensions, if each corner costs the same as $\alpha = \frac{1}{4}$ length of perimeter, then the "best" shape is a square.  The same problem in higher dimensions, for arbitrary $\alpha$, may be difficult.  In fact, in higher dimensions, for small $\alpha$, the answer is not even a regular solid!
