Maximum volume cylinder from a sheet of paper 
We have an A4 sheet of paper and we have to build a cylinder of maximum volume using this paper by cutting out  a rectangle and the two base circles in order to make the construction. 

My approach: denote $\ell$ and $L$ the dimensions of the sheet of paper, $r$ the radius of the cylinder and $h$ the height of the cylinder.
I can work  with determining the maximum,(using derivatives after i obtain the formula of the volume), but the problem is : how do i precisely "model" this situation, as to be sure i obtain the maximum volume?
I see two cases, regarding where i "put" the two circles: "along" the shortest or the longest side of the paper, buuuut... it seems that it would remain some portion of the paper unused...(not only when i cut out the circles, which is inevitable, but even at the rectangle which would be wrapped in order to build the cylinder). 
I can compute a "theoretical" maximum, assuming that i could use ALL the surface of the sheet, but that cylinder couldn;t be practically built, or i don;t see how. So this theoretical max is not attainable, but my max...i'm not sure it's the real max. If you get what i mean...
 A: A4 paper measures 210*297mm or 595 X 842 points. For maximum volume we want height and diameter of the cylinder to be equal. The circle(s) diameter will be $d$ and the rectangle will be $dX\pi d$. Suppose we take one circle off the end for now–perhaps we can get two. Then we have two scenarios. 
If we place the rectangle lengthwise, we have $d+\pi d=842$. So $d=\frac{842}{1+3.1415926}=203.3$. Since the paper is $595$ points wide, we can get $2$ circles 203.3 points off the end and $2$ rectangles from the main body, each $203.3X638.7$ points.
If we place the rectangle across the paper, we have a max of 595 points or $189.4$ points for $d$.
Suppose, however, that we make the rectangle out of two smaller rectangles. Each diameter and rectangle width would be half the paper width wide: $\frac{595}{2}=297.5$ points. The length of the rectangle after the two halves are joined, end-to-end, would be $297.5*3.1415926=934.6$.
Half this length would be $457.3$ and that would leave $842-457.3-297.5=87.2$ points of waste on the end of the paper. Not bad. Hope this helps.
