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From: David Entwistle <qnivq.ragjvfgyr@ogvagrearg.pbz>
Newsgroups: rec.puzzles
Subject: Re: How to Make Cisterns
Date: Wed, 7 May 2025 09:25:15 -0000 (UTC)
Organization: A noiseless patient Spider
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On Sat, 3 May 2025 08:08:42 -0000 (UTC), David Entwistle wrote:

> The puzzle is what size should the cut out pieces be, such that the
> cistern will hold the greatest possible quantity of water?

ANSWER
NSWER
SWER
WER
ER
R

I hope this is correct - it's been a while since I did any mathematics. It 
is good to have a refresher...

As the square sheet of zinc has side length one metre, then, if the length 
of each of of the cut-outs is h, then the dimensions of the final cistern 
are (1 - 2h), (1 - 2h) and h. The volume (V) is the product of dimensions. 
Multiplying out we get:

V = 4h^3 - 4h^2 + h + 0

To find the maximum volume we can differentiate and set dV/dh to zero. 
These are the points where volume does not change with a change in the 
value of h. These may be maxima, minima, or points of inflection.

dV/dh = 12h^2 - 8h + 1

Setting dV/dh to zero gives:

12h^2 -8h + 1 = 0

Factorizing:

(6h - 1)(2h - 1) = 0

So dV/dh = 0 when h = 1/6, or when h = 1/2.

We can determine whether each point is the maxima, minima or point of 
inflection by examining the curve, looking at points either side of the 
point, or using the second derivative. h = 1/6 (0.1667 m) is the local 
maxima in the range 0 < h< 1/2. The volume at that point is given by:

V = 4(1/6)^3 - 4(1/6)^2 + 1/6

This simplifies to:

V = 2/27

That's roughly 0.074 cubic metres.

-- 
David Entwistle