Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP
Posting-Version: version B 2.10.1 6/24/83; site dciem.UUCP
Path: utzoo!dciem!ntt
From: ntt@dciem.UUCP (ntt)
Newsgroups: net.math
Subject: Answer to: Everything you know is wrong!
Message-ID: <286@dciem.UUCP>
Date: Tue, 16-Aug-83 16:37:02 EDT
Article-I.D.: dciem.286
Posted: Tue Aug 16 16:37:02 1983
Date-Received: Tue, 16-Aug-83 20:01:40 EDT
References: <557@ihuxr.UUCP>
Organization: D.C.I.E.M, Toronto, Canada
Lines: 18

I assume the item referred to was intended as a joke or puzzle!
Certainly, it is true that for the objects specified we have:

     sphere:     volume = (4/3)*pi*r^3       surface = 4*pi*r^2
     cylinder:   volume = (6/3)*pi*r^3       surface = 6*pi*r^2
since the cylinder has h = 2*r.

And it is true that the surface/volume ratios are numerically equal to 3/r.
But the objects being compared are not equal in volume, nor in area (if r>0).
The standard claim about the sphere having the smallest surface/volume
ratio refers to comparisons among objects of equal volume (or of equal
surface).  To equate the volumes in this case we must see that r has
different meanings for the sphere and the cylinder, and force
      r[sph]/r[cyl] = (3/2)^(1/3)
And then 3/r[sph] is not equal to 3/r[cyl].
Alternatively we can equate surface areas, in which case 1/2 replaces 1/3.

Mark Brader, NTT Systems Inc., Toronto (decvax!dciem!ntt)