28 views

Skip to first unread message

Dec 31, 2001, 11:45:44 PM12/31/01

to

"The Free Plot" (which appeared here recently) is a very nice

two-dimensional puzzle. Here's an analogous three-dimensional puzzle:

Within a given cube, what solid has the smallest ratio of surface

area to volume, and what is that ratio?

two-dimensional puzzle. Here's an analogous three-dimensional puzzle:

Within a given cube, what solid has the smallest ratio of surface

area to volume, and what is that ratio?

I doubt that I have the answer, but I can give a value (an upper bound)

which is probably rather close to the minimum ratio. For those who wish

to work on this problem before seeing what I've done so far,

S

P

O

I

L

E

R

S

P

A

C

E

First, note that, just as in "The Free Plot", many people will tend to

jump to wrong conclusions, thinking that either the given cube itself

or the sphere inscribed therein will have the minimum ratio of surface area

to volume. In fact, the cube and the sphere have the same ratio, 6/e, where

e denotes the edge length of the cube. Henceforth, let's assume that we are

dealing with a cube of unit edge length, so the surface area to volume

ratio for both the cube and the sphere is 6. But this is certainly not the

minimum ratio obtainable. A solid produced by rounding the edges and

corners of the cube somewhat can have a smaller ratio.

Proceeding by analogy with the solution to "The Free Plot", I considered

a solid formed from the given cube by rounding all of its edges, leaving

them "quarter round" with radius r. Thus, any orthogonal projection

of the solid onto an original cube face would resemble the solution to

"The Free Plot" (although the radius r of the quarter circles would not

necessarily be 1/(2 + Sqrt(pi)), of course). Also consider the parts of

the solid contained within small cubes of edge length r nestled in the

corners of the original cube. These eight parts could be assembled to form

a tricylinder (a Steinmetz solid, the intersection of three cylinders). It

is only on these parts that the solid has edges or corners. They are rather

blunt, but surely the fact that the solid has them at all tells us that a

still smaller ratio of surface area to volume can be obtained, by using a

smooth solid. [Unfortunately, I don't know precisely what that smooth solid

should be. It might be interesting to investigate solids given by

|x|^p + |y|^p +|z|^p <= 1/2^p for different values of p.]

For the type of solid I investigated, obtained by "quarter rounding" of

the cube of unit edge length,

surface area A = 6(1-2r)^2 + 6(pi)r(1-2r) + 24(2-Sqrt(2))r^2 and

volume V = (1-2r)^3 + 6r(1-2r)^2+ 3(pi)r^2(1-2r) + 8(2-Sqrt(2))r^3.

Minimizing A/V for such a solid is straightforward, leading to the

solution of a quartic equation. Rather than giving a very messy precise

expression for the value of r at which the minimum is attained, let me

just say that it is approximately r = 0.2376425, at which A/V is

approximately 5.45525 .

As noted previously, surely this is not the best possible, due to the

corners and edges of this type of solid. I have no idea whether finding

the solid giving minimum A/V will be difficult or not, but it seems like

an interesting problem.

Happy New Year!

David Cantrell

--

-------------------- http://NewsReader.Com/ --------------------

Usenet Newsgroup Service

Jan 1, 2002, 7:10:57 AM1/1/02

to

http://groups.google.com/groups?selm=38F7980F.AD663055%40pisquaredoversix.

force9.co.uk

force9.co.uk

may be apposite.

Wilma.

Reply all

Reply to author

Forward

0 new messages

Search

Clear search

Close search

Google apps

Main menu