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From: brent@itm.UUCP
Newsgroups: net.math
Subject: Area to Volume ratios
Message-ID: <1029@itm.UUCP>
Date: Tue, 16-Aug-83 15:59:37 EDT
Article-I.D.: itm.1029
Posted: Tue Aug 16 15:59:37 1983
Date-Received: Wed, 17-Aug-83 18:57:07 EDT
Lines: 25


As Lew pointed out, a sphere with radius r and a cylinder with radius
r and height 2r have the same surface area to volume ratio: 3/r
Now try a cube of edge length 2r.  6*(2*r)^2 / (2*r)^3 = 3/r
The same as the cube and the cylinder.  What's wrong?

The ratio of surface area to volume depends on what you call r.
Example: try a unit cube with edge length r.  The S/A ratio is
6/r.  Compare this with the 3/r answer obtained above.

The proper phrasing of the constraint is something like "The sphere
gives the minimum surface area of any shape *for any given volume*"
Compute the surface areas for a cube, a cylinder and a sphere
of volume 1.  It comes out something like:

        Shape      Volume       Area
        Cube        1            6
        Cylinder    1            5.54
        Sphere      1            4.84

Indeed the sphere is the minimum area for the given volume enclosed.
The previous results of 3/r resulted from clever choices for r,
thus comparing the ratios of shapes enclosing different volumes.

            Brent Laminack  (msdc!itm!brent)