Make a paper can with the greatest volume possible by cuttting and
taping together two circles and a rectangle form 8 1/2 by 11 inch
paper. It only can be one sheet of paper. What are the circumferences,
diameters, or radiusis of the circles? What is the length and width of
the rectangle?

+-------+   cylinder circumference = 8.5 in.
  |      O|     -> endcap diameter = 8.5/pi = 2.706 in.
  |      O|     -> cylinder height = 11 - 2.706 = 8.294 in.
  +-------+     -> endcap area = pi * (2.706/2)^2 = 5.751 sq.in.
                -> cylinder volume = 8.294 * 5.751 = 47.699 cu.in.

  +-------+   cylinder circumference = 11 in.
  |OO     |     -> endcap diameter = 11/pi = 3.501 in.
  |       |     -> cylinder height = 8.5 - 3.501 = 4.999 in.
  +-------+     -> endcap area = pi * (3.501)^2 = 9.627 sq.in.
                -> cylinder volume = 4.999 * 9.627 = 48.125 cu.in.

  +-------+   So cutting the paper this way yields the cylinder
  |OO     |   with the largest volume.
  |       |
  +-------+

Thank you!

Do you have an explaination in words on how you got your answer?

If I understand the question,willcodeforfood neglected to get the
material for the end disks from the same peice of paper that the
rectangle comes from. This reduces the maximumum volume considerably
and makes the problem very complex, unless the scrap from the disks
can be pasted on to the rectangle in an infinate number of patches.
 Assuming the scrap will not be utilized, my guess is trial and error
is the most practical solution.   Neil

I think you are wrong, Neil, but I have an additonal suggestion.

Willco's answer seems correct, IF the encaps must be just one piece of
paper each, which most people would assume, but this was not stated in
the question, nor probably reguired by whomever proposed the project. 
If you make the endcaps each from two semicircles, the height of the
cylinder can be increased by 1.35 in.
Probably one can use even more of the wasted paper by using strips to
make the endcaps, but then it gets tricky - not just if you actually
have to make your paper can.
If this were allowed, the mathematical absolute maximum could be real
tricky, for me at least, and practically impossible.  It would be a
calculus problem -
no maybe it wouldn-t be, NO  - I'm trying to think.  
Just calculate the max. volume cylinder with endcaps that would have a
surface area of 8.5 by 11 inches.  (maybe that does require calculus,
don't know)
The result practically would be that the endcaps were a mass of
infinitely thin threads of paper laid side by side to create circles.

I'd stick with semicircles or maybe only four pieces to make the
endcaps and just the suggest the ultimate solution  -
All presupposing that there was no rule saying the endcaps had to be
single pieces of paper.
But you would still "wow" the "project leader" with the recognition of
the theoretical max. solution.

This problem does not allow the trickery suggested above.  The problem
statement clearly asks for two circles and one rectangle, not four
semicircles and a rectangle, or a bunch of little strips of paper.

That having been said, I thought I'd calculate the theoretical maximum
volume given a loosening of the problem's constraints as suggested by
others.  The difficulty of calculating the theoretical maximum makes
it a precalculus question, albeit a fairly tough one.

===========================================

Given:
  The circles each have radius r
  The rectangle has length l and height h
  Pi will be represented by p

The length of the rectangle must be equal to the circle's circumference
  ->  l = 2pr

The total area of the rectangle and two circles is 8.5 * 11
  ->  lh + 2pr^2 = 93.5

The volume of the cylinder V = base * height
  ->  V = pr^2 * h

===========================================

We start by determining h in terms of r

  lh + 2pr^2 = 93.5
  2pr(h + r) = 93.5
      h + r  = 93.5/2pr         (r!=0, but we knew that already)
           h = 93.5/2pr - r

Next we calculate V in terms of r

  V = pr^2 * h
  V = pr^2 * (93.5/2pr - r)
  V = 93.5pr^2/2pr - pr^3
  V = 93.5r/2 - pr^3

We know dV (V') approaches zero as r approaches a value with max V

  V' = 93.5/2 - 3pr^2

Now solving for V' = 0

  0 =  = 93.5/2 - 3pr^2
  3pr^2 = 93.5/2
  r = square root of: 93.5/2*3*(3.1415)
  r = 2.2272

The radius of each circle:         2.2272 in.
The area of each circle:           15.5836 sq.in.
The circumference of each circle:  13.9938 in.
The length of the rectangle:       13.9938 in.   (paper length = 11.5 in. !!!)
We can solve now for h...
The height of the rectangle:       4.4543 in.

The maximum theoretical volume is thus: 69.4140 cu.in.

===========================================

neilzero-ga, Try cutting a piece of paper according to the dimensions
I first posted.  You will find that all three shapes come from a
single sheet of paper.  Trial and error is almost never the most
practical way to solve an optimization and one can never assert an
optimum solution after using such a casual approach.

Hi willcodeforfood: I don't know how you determined that the problem
does not allow trickery, such as patching some of the scrap onto the
end of the rectangle to get the 13.9938 inches which you got for the
length of the rectangle and the circumfrence of the cylinder. The area
of paper you used totals about 72.75 square inches while the 8.5 by 11
paper has an area of 93.5 square inches, so there is 20.75 square
inches of unused scrap.
 I applied the engineering method to determine if your numbers were
possible. From a piece of paper 8.5 by 12.5; the longest rectangle
possible had a length of 11.75 inches with a width of 4.4543 (that is
cut diagonally from the paper, which reduces the length of the
rectangle) The scrap is not large enough to cut a 2.2272 radius circle
in one piece, however two circle can be made in two installments each
from the scrap, with sufficient scrap left to extend the rectangle to
13.8838 inches or somewhat more, even if the paper is 8.5 by 11
inches.
We can get a circumfrence of 11 inches from 11 inches of paper, and
make two 2.02 inch radius circles out of the scrap, but not 2.2272
inch radius circles except in two installments each.  Neil

Hi neil,

As I see it, the problem explicitly calls for cutting two circles and
a rectangle from a single sheet, not cutting the paper up into any
number of shapes and then pasting them together to form a cylinder. 
If you see the wording as allowing the use of scrap to patch your can
together, I don't agree, but I accept that you see it that way.

=========================================================

In my first post, I suggested the maximum practical volume one could
achieve was 48.125 cu.in. and you suggested that my solution
"neglected to get the material for the end disks."  To this I
suggested you construct a can following this pattern:

  +-------+   cylinder circumference = 11 in.
  |OO     |     -> endcap diameter = 11/pi = 3.501 in.
  |       |     -> cylinder height = 8.5 - 3.501 = 4.999 in.
  +-------+     -> endcap area = pi * (3.501)^2 = 9.627 sq.in.
                -> cylinder volume = 4.999 * 9.627 = 48.125 cu.in.

1. Cut the paper lengthwise into two strips, with the two 
   strips having widths 3.501 in. and 4.999 in. respectively.
2. Wrap the wider strip into a tube and tape it.
3. From the narrower, unused strip cut two circles, each
   with diameter of 3.501 in.
4. Tape one circle on each end of the tube.
5. Throw away the unusable 3.998 in. * 3.501 in.scrap.

=========================================================

In my subsequent post, I calculated a "theoretical maximum" based on
"a loosening of the problem's constraints."  By this I meant that I
would determine the dimensions of a can with maximum volume having a
total surface area the same as a 8.5 x 11 inch sheet of paper.  Thus
you see the use of derivatives and polynomial expressions.  Obviously
one could not construct this theoretical can from a real sheet of
paper, especially given that it uses a rectangle with a length of
almost 14 inches, well beyond the length of the original piece of
paper.

The theoretical can is composed of three shapes:

  rectangle:       area = 4.4543 * 13.9938 = 66.3326 sq.in.
  circle (top):    area =                    15.5836 sq.in.
  circle (bottom): area =                    15.5836 sq.in.
                                             --------------
  total area of all shapes:                  93.5000 sq.in.

  total area of original sheet of paper:
  8.5 in. * 11 in. =                         93.5000 sq.in.

I'm at a loss trying to follow your engineering method or figure out
what you mean when you say that "there is 20.75 square inches of
unused scrap" in the theoretical solution I proposed.  Perhaps you may
have calculated incorrectly.  If you feel there is a more optimum
solution to the practical or theoretical approach, by all means
suggest it and show your work.  If you see an error in my logic or
calculations (and there may well be such), then please explain where
I've gone awry.  Stating that my solution is off by X does not seem
particularly helpful.

Oh, you're right, Willco, on both solutions, and your second indicates
that if my breaking the rules were allowed, even on a lower, manually
possible level, it is better to "patch" the rectangle for the tube,
which would be much easier.
I still think the description of the project lays the emphasis on
creating the maximum volume, only describing that the round can with
top and bottom must be created from one sheet of 8.5 by 11 inch paper;
and that the "teacher" probably didn't expect more than your first
solution, but certainly wouldn't be upset by one the went beyond his
own expectations.  And Ikatz could alway provide that and then expand
on it  - to prove how good he is at using GA  :-)

Ikatz:  maybe you could come back later and tell us the result ...  Please!