tag:blogger.com,1999:blog-2433841880619171855.post742361390070562116..comments2024-03-27T21:09:44.320+00:00Comments on Pat'sBlog: Repeat of an Isoperimetric DiscoveryUnknownnoreply@blogger.comBlogger2125tag:blogger.com,1999:blog-2433841880619171855.post-62718407885134530452011-02-13T16:53:16.484+00:002011-02-13T16:53:16.484+00:00James,
Thanks, that may just be it. I'm not ...James,<br /> Thanks, that may just be it. I'm not sure that will ease the burden of a general proof, but it seems to make the whole idea reasonable.Pat's Bloghttps://www.blogger.com/profile/15234744401613958081noreply@blogger.comtag:blogger.com,1999:blog-2433841880619171855.post-11777103715273741582011-02-13T16:08:50.665+00:002011-02-13T16:08:50.665+00:00The Arithmetic-Geometric Mean Inequality is at the...The Arithmetic-Geometric Mean Inequality is at the heart of this I think. For example, in the first picture you draw, let the large rectangle be w by h. Then the constraint is w + 3h = P, the given total length of fence. By the AMGM inequality<br /><br />P/2 = (w + 3h)/2 is greater than or equal to sqrt(w 3h) = sqrt(3A) where A is the total area. Therefore<br /><br />A is less than or equal to P^2/12 with equality (and this is the crucial point) when w = 3h, i.e. precisely when the amount of fence used in each direction is the same.Anonymoushttps://www.blogger.com/profile/04032324922719374790noreply@blogger.com