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Maximizing the Volume of a Box


Date: 06/27/99 at 17:16:20
From: Michael Sherrillo
Subject: Geometry

I have a piece of glass that is 14 inches wide and 72 inches long. I 
need to build glass cage. What dimensions would I need to make the 
cage to get the maximum volume? What dimensions would I need if I only 
had to make 5 sides instead of six? This is a real life problem I have 
right now, not a homework problem.

Sincerely,
Michael Sherrillo


Date: 06/28/99 at 12:18:38
From: Doctor Peterson
Subject: Re: Geometry

Hi, Michael.

This does sound a lot like a homework problem you'd get in calculus, 
but the addition that you have to make it from one long narrow piece 
of glass makes it a little trickier, since there are lots of ways to 
cut the pieces from the sheet. At the same time, since it's a real 
life problem, we don't have to solve it precisely!

Without restrictions on the shape of the materials, the general 
solution for your first problem (maximum volume for a closed box with 
fixed area) is to make a cube, and for your second problem (for an 
open box) is to cut that cube in half (the height being half the 
dimensions of the open top).

If we look at how we can cut your glass to something close to those 
dimensions, we can get a pretty good solution by doing the obvious:

                         72
    +--------+--------+------+------+------+------+
    |        |        |      |      |      |      |
    |a       |a       |a     |a     |b     |b     |14
    |        |        |      |      |      |      |
    |   b    |   b    |  c   |  c   |  c   |  c   |
    +--------+--------+------+------+------+------+

We want a = b = 14 and 2b + 4c = 72, which gives c = 11. That's 
probably close enough to a cube to be about the best you can do. 
The volume will be 14*14*11 = 2156 in^3, compared to the theoretical 
maximum (with all three sides sqrt(14*72/6) = 12.96 in) of 2177.5 
in^3. Not bad!

With an open top, we have to leave out one b*c side. If we try the 
same layout,

                      72
    +--------+--------+------+------+------+
    |        |        |      |      |      |
    |a       |a       |a     |a     |b     |14
    |        |        |      |      |      |
    |   b    |   b    |  c   |  c   |  c   |
    +--------+--------+------+------+------+

we get a = b = 14 and 2b+3c = 72, which gives c = 14.66 in. Since c 
ought to be half of a, or 7, this seems pretty far off. In this case 
the theoretical maximum, with a square bottom sqrt(14*72/3) = 18.33 
in, is 3079 in^3; our dimensions would give 14*14*14.66 = 2874.66, 
which is indeed much smaller than we would hope for. On the other 
hand, since we can't have the square bottom bigger than 14 inches, we 
probably can't do any better. (I'm assuming you can't put together the 
bottom from two glass panels glued side by side.) 

I think either of these answers should be good enough. Let me know if 
you need more help.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Geometry
High School Polyhedra
High School Practical Geometry

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