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From: Mike Terry <news.dead.person.stones@darjeeling.plus.com>
Newsgroups: rec.puzzles,sci.math
Subject: Re: Maximize Cistern Volume -- (cut out 4 squares (at Corners) and
 discard them)
Date: Tue, 6 May 2025 23:29:51 +0100
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On 06/05/2025 20:47, Richard Tobin wrote:
> In article <14b4afbbf6091c2c839beec0c3c41f21@www.novabbs.com>,
> HenHanna  <HenHanna@dev.null> wrote:
> 
>> What's not at all obvious (intuitive) for me is.... why or how
>>                        the max Volume is achieved at   x=1/6
> 
> Note that x=1/6 makes the total area of the sides equal to the area of
> the base (4/9).  I wouldn't be surprised if that is a special case of
> some more general result.
> 
> -- Richard
> 

That makes sense - when we make the 4 cutout squares bigger, increasing their side length by a very 
small amount s, the effect on the cistern is, broadly
a)  increase the height by s, which /increases/ its volume by A.s
     [A being the area of the base]
b)  decrease the "radius" of the box by s, which /decreases/ its volume by B.s
     [B being the area of the sides]
So at the maximum volume these two effects must cancel out, and we will have A = B.  Yes there are 
higher order changes in volume involving s^2 and higher powers, but we neglect those as small 
compared to first order changes.

This is in effect doing calculus from scratch, ignoring higher order terms in s to get the 
derivative dV/dx which is set to zero.  The ignoring of higher terms is like what happens in the 
proof of the product rule for derivatives.


Mike.