Sujet : Re: How to Make Cisterns
De : qnivq.ragjvfgyr (at) *nospam* ogvagrearg.pbz (David Entwistle)
Groupes : rec.puzzlesDate : 07. May 2025, 10:25:15
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <vvf8tr$t9bn$2@dont-email.me>
References : 1
User-Agent : Pan/0.149 (Bellevue; 4c157ba git@gitlab.gnome.org:GNOME/pan.git)
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
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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