Adapted from "Amusements in Mathematics" by Henry Ernest Dudeney.

Given a large sheet of zinc, measuring (before cutting) one metre square,
you are asked to cut out square pieces (all of the same size) from the
four corners of this sheet. You are then to fold up the sides of the
resulting shape, solder the edges and make a cistern. The cistern is open
at the top. You can assume you have the appropriate tools and skill to
carry out the task.

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

I guess a follow on question could be; is it possible to cut out four non- square pieces (all of the same size and shape), producing a cistern with tapering sides, which has greater volume. I haven't looked at that yet...
--
David Entwistle

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On 03/05/2025 09:08, David Entwistle wrote:

Adapted from "Amusements in Mathematics" by Henry Ernest Dudeney.

Given a large sheet of zinc, measuring (before cutting) one metre square,
you are asked to cut out square pieces (all of the same size) from the
four corners of this sheet. You are then to fold up the sides of the resulting shape, solder the edges and make a cistern. The cistern is open
at the top. You can assume you have the appropriate tools and skill to
carry out the task.

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

.....I cheated, and after this spoiler I offer an answer to the
nearest micron.
.....I cheated, and after this spoiler I offer an answer to the
nearest micron
.....I cheated, and after this spoiler I offer an answer to the
nearest micro
.....I cheated, and after this spoiler I offer an answer to the
nearest micr
.....I cheated, and after this spoiler I offer an answer to the
nearest mic
.....I cheated, and after this spoiler I offer an answer to the
nearest mi
.....I cheated, and after this spoiler I offer an answer to the
nearest m
.....I cheated, and after this spoiler I offer an answer to the
nearest
.....I cheated, and after this spoiler I offer an answer to the
nearest
.....I cheated, and after this spoiler I offer an answer to the
neares
.....I cheated, and after this spoiler I offer an answer to the neare
.....I cheated, and after this spoiler I offer an answer to the near
.....I cheated, and after this spoiler I offer an answer to the nea
.....I cheated, and after this spoiler I offer an answer to the ne
.....I cheated, and after this spoiler I offer an answer to the n
.....I cheated, and after this spoiler I offer an answer to the
.....I cheated, and after this spoiler I offer an answer to the
.....I cheated, and after this spoiler I offer an answer to th
.....I cheated, and after this spoiler I offer an answer to t
.....I cheated, and after this spoiler I offer an answer to
.....I cheated, and after this spoiler I offer an answer to
.....I cheated, and after this spoiler I offer an answer t
.....I cheated, and after this spoiler I offer an answer
.....I cheated, and after this spoiler I offer an answer
.....I cheated, and after this spoiler I offer an answe
.....I cheated, and after this spoiler I offer an answ
.....I cheated, and after this spoiler I offer an ans
.....I cheated, and after this spoiler I offer an an
.....I cheated, and after this spoiler I offer an a
.....I cheated, and after this spoiler I offer an
.....I cheated, and after this spoiler I offer an
.....I cheated, and after this spoiler I offer a
.....I cheated, and after this spoiler I offer
.....I cheated, and after this spoiler I offer
.....I cheated, and after this spoiler I offe
.....I cheated, and after this spoiler I off
.....I cheated, and after this spoiler I of
.....I cheated, and after this spoiler I o
.....I cheated, and after this spoiler I
.....I cheated, and after this spoiler I
.....I cheated, and after this spoiler
.....I cheated, and after this spoiler
.....I cheated, and after this spoile
.....I cheated, and after this spoil
.....I cheated, and after this spoi
.....I cheated, and after this spo
.....I cheated, and after this sp
.....I cheated, and after this s
.....I cheated, and after this
.....I cheated, and after this
.....I cheated, and after thi
.....I cheated, and after th
.....I cheated, and after t
.....I cheated, and after
.....I cheated, and after
.....I cheated, and afte
.....I cheated, and aft
.....I cheated, and af
.....I cheated, and a
.....I cheated, and
.....I cheated, and
.....I cheated, an
.....I cheated, a
.....I cheated,
.....I cheated,
.....I cheated
.....I cheate
.....I cheat
.....I chea
.....I che
.....I ch
.....I c
.....I
.....I
.....
....
...
..
.

Side = 999999 microns, volume = 698521279 cubic microns.
Side = 999997 microns, volume = 791833765 cubic microns.
Side = 999995 microns, volume = 909146179 cubic microns.
Side = 999993 microns, volume = 1050458473 cubic microns.
Side = 999991 microns, volume = 1215770599 cubic microns.
Side = 999989 microns, volume = 1405082509 cubic microns.
Side = 999987 microns, volume = 1618394155 cubic microns.
Side = 999985 microns, volume = 1855705489 cubic microns.
Side = 999983 microns, volume = 2117016463 cubic microns.
Side = 999819 microns, volume = 2131502515 cubic microns.
Side = 998659 microns, volume = 2137812763 cubic microns.
Side = 998374 microns, volume = 2144275288 cubic microns.
Side = 997379 microns, volume = 2145052699 cubic microns.
Side = 995252 microns, volume = 2147110208 cubic microns.
Side = 930405 microns, volume = 2147409565 cubic microns.
Side = 925780 microns, volume = 2147420992 cubic microns.
Side = 912006 microns, volume = 2147446488 cubic microns.
Side = 777683 microns, volume = 2147468811 cubic microns.
Side = 461441 microns, volume = 2147475329 cubic microns.
Side = 357737 microns, volume = 2147481849 cubic microns.

The last entry is of course the winner.

I guess a follow on question could be; is it possible to cut out four non- square pieces (all of the same size and shape), producing a cistern with tapering sides, which has greater volume. I haven't looked at that yet...

"Easily" was my immediate reaction... but then I thought about
it, and decided that it... well, that it's a far better question
than it looks. :-)

--
Richard Heathfield
Email: rjh at cpax dot org dot uk
"Usenet is a strange place" - dmr 29 July 1999
Sig line 4 vacant - apply within

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On Sat, 3 May 2025 11:07:57 +0100, Richard Heathfield wrote:

.....I cheated, and after this spoiler I offer an answer to the nearest micron.

In your last couple of entries, approaching an answer, the side changes by
30% and the volume changes by very little (less than 0.001%). I think
something as gone off track somewhere.

If you can recall high-school calculus, then the answer falls out as a
simple fraction.

"Easily" was my immediate reaction... but then I thought about it, and decided that it... well, that it's a far better question than it looks.

My initial reaction was "no", but then I began to have doubts. I'll give
it a go.


--
David Entwistle

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

In article <vv4ptt$399aa$1@dont-email.me>,
Richard Heathfield <rjh@cpax.org.uk> wrote:

.....I cheated

Using 32-bit ints?

-- Richard

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On 03/05/2025 13:19, David Entwistle wrote:

On Sat, 3 May 2025 11:07:57 +0100, Richard Heathfield wrote:

.....I cheated, and after this spoiler I offer an answer to the nearest
micron.

In your last couple of entries, approaching an answer, the side changes by 30% and the volume changes by very little (less than 0.001%). I think something as gone off track somewhere.

Nope. Rather than post a million results, I only posted the
incremental improvements:

include <stdio.h>

int main(void)
{
int i;
int ms = 0;
int mv = 0;
for(i = 0; i < 1000000; i++)
{
int s = 1000000 - i;
int v = s * s * s;
if(v > mv)
{
ms = s;
mv = v;
printf("Side = %d microns, volume = %d cubic microns.\n",
s, v);
}

}
return 0;
}

[Aside to Richard Tobin - yes, I should have used long ints. Sue me.]

If you can recall high-school calculus, then the answer falls out as a
simple fraction.

I can recall passing the exam, and the answer falls out
as...um... 1000000-357737 remaining=642263 removed, so the
squares that were removed were each 642263/2=321131 microns on a
side.

The simple fraction is therefore

321131
-------
1000000

but that's probably not what you meant.

"Easily" was my immediate reaction... but then I thought about it, and
decided that it... well, that it's a far better question than it looks.

My initial reaction was "no", but then I began to have doubts. I'll give
it a go.

Are we allowed to use offcuts to patch holes? I'm not saying I
plan to try, but it strikes me that it would do no harm to ensure
that the ground rules are the same for everybody.

--
Richard Heathfield
Email: rjh at cpax dot org dot uk
"Usenet is a strange place" - dmr 29 July 1999
Sig line 4 vacant - apply within

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

In article <vv4iua$31980$1@dont-email.me>,
David Entwistle <qnivq.ragjvfgyr@ogvagrearg.pbz> wrote:

I guess a follow on question could be; is it possible to cut out four non- >square pieces (all of the same size and shape), producing a cistern with >tapering sides, which has greater volume. I haven't looked at that yet...

Yes, because if you open up the top of the cistern by a small amount
(turning the cut-off squares into kites) it will obviously increase
the mean horizontal cross-section more than it will decrease the
height.

To prove that this is actually, rather then merely obviously, true,
and to find the maximum, would require more manipulation than I am
currently willing to do by hand.

-- Richard

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

In article <vv55q1$3jq94$1@dont-email.me>,
Richard Heathfield <rjh@cpax.org.uk> wrote:

int main(void)
{
int i;
int ms = 0;
int mv = 0;
for(i = 0; i < 1000000; i++)
{
int s = 1000000 - i;
int v = s * s * s;

Oh no it isn't. The cistern is not a cube.

I'm not sure which side your s is. If s is the side of the cistern,
then the height h is (1000000-s)/2, and v = s * s * h. If s is the
side of the cut-off square, then the base b is (1000000-2*s) and
v = b * b * s.

And of course the calculation overflows 32-bit ints.

-- Richard

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On Sat, 3 May 2025 15:43:20 +0100, Richard Heathfield wrote:

.....Looks like I need a new spoiler (and a brain transplant).

Well done. You can put off that transplant for a while.

The algebraic solution is relatively straight-forward. I'll leave it a bit
and then post it.

--
David Entwistle

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On 03/05/2025 14:52, Richard Tobin wrote:

In article <vv55q1$3jq94$1@dont-email.me>,
Richard Heathfield <rjh@cpax.org.uk> wrote:

int main(void)
{
int i;
int ms = 0;
int mv = 0;
for(i = 0; i < 1000000; i++)
{
int s = 1000000 - i;
int v = s * s * s;

Oh no it isn't. The cistern is not a cube.

Bugger 1.

I'm not sure which side your s is. If s is the side of the cistern,
then the height h is (1000000-s)/2, and v = s * s * h. If s is the
side of the cut-off square, then the base b is (1000000-2*s) and
v = b * b * s.

And of course the calculation overflows 32-bit ints.

Bugger 2!

Redoubled in buggers. I quit!

(For a few moments, at least.)

--
Richard Heathfield
Email: rjh at cpax dot org dot uk
"Usenet is a strange place" - dmr 29 July 1999
Sig line 4 vacant - apply within

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On 03/05/2025 15:26, Richard Heathfield wrote:

<snip>

Redoubled in buggers. I quit!

(For a few moments, at least.)

.....Looks like I need a new spoiler (and a brain transplant).
.....Looks like I need a new spoiler (and a brain transplant)
.....Looks like I need a new spoiler (and a brain transplant
.....Looks like I need a new spoiler (and a brain transplan
.....Looks like I need a new spoiler (and a brain transpla
.....Looks like I need a new spoiler (and a brain transpl
.....Looks like I need a new spoiler (and a brain transp
.....Looks like I need a new spoiler (and a brain trans
.....Looks like I need a new spoiler (and a brain tran
.....Looks like I need a new spoiler (and a brain tra
.....Looks like I need a new spoiler (and a brain tr
.....Looks like I need a new spoiler (and a brain t
.....Looks like I need a new spoiler (and a brain
.....Looks like I need a new spoiler (and a brain
.....Looks like I need a new spoiler (and a brai
.....Looks like I need a new spoiler (and a bra
.....Looks like I need a new spoiler (and a br
.....Looks like I need a new spoiler (and a b
.....Looks like I need a new spoiler (and a
.....Looks like I need a new spoiler (and a
.....Looks like I need a new spoiler (and
.....Looks like I need a new spoiler (and
.....Looks like I need a new spoiler (an
.....Looks like I need a new spoiler (a
.....Looks like I need a new spoiler (
.....Looks like I need a new spoiler
.....Looks like I need a new spoiler
.....Looks like I need a new spoile
.....Looks like I need a new spoil
.....Looks like I need a new spoi
.....Looks like I need a new spo
.....Looks like I need a new sp
.....Looks like I need a new s
.....Looks like I need a new
.....Looks like I need a new
.....Looks like I need a ne
.....Looks like I need a n
.....Looks like I need a
.....Looks like I need a
.....Looks like I need
.....Looks like I need
.....Looks like I nee
.....Looks like I ne
.....Looks like I n
.....Looks like I
.....Looks like I
.....Looks like
.....Looks like
.....Looks lik
.....Looks li
.....Looks l
.....Looks
.....Looks
.....Look
.....Loo
.....Lo
.....L
.....
....
...
..
.

Height = 166667.000000, side = 666666.000000 microns, volume = 74074074073851856.000000 cubic microns.

So 1/6, which is a suspiciously neat fraction.

--
Richard Heathfield
Email: rjh at cpax dot org dot uk
"Usenet is a strange place" - dmr 29 July 1999
Sig line 4 vacant - apply within

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

In article <vv5a28$3n6ot$2@dont-email.me>,
Richard Heathfield <rjh@cpax.org.uk> wrote:

On 03/05/2025 15:26, Richard Heathfield wrote:

<snip>

Redoubled in buggers. I quit!

(For a few moments, at least.)

.....Looks like I need a new spoiler (and a brain transplant).
.....Looks like I need a new spoiler (and a brain transplant)
.....Looks like I need a new spoiler (and a brain transplant
.....Looks like I need a new spoiler (and a brain transplan
.....Looks like I need a new spoiler (and a brain transpla
.....Looks like I need a new spoiler (and a brain transpl
.....Looks like I need a new spoiler (and a brain transp
.....Looks like I need a new spoiler (and a brain trans
.....Looks like I need a new spoiler (and a brain tran
.....Looks like I need a new spoiler (and a brain tra
.....Looks like I need a new spoiler (and a brain tr
.....Looks like I need a new spoiler (and a brain t
.....Looks like I need a new spoiler (and a brain
.....Looks like I need a new spoiler (and a brain
.....Looks like I need a new spoiler (and a brai
.....Looks like I need a new spoiler (and a bra
.....Looks like I need a new spoiler (and a br
.....Looks like I need a new spoiler (and a b
.....Looks like I need a new spoiler (and a
.....Looks like I need a new spoiler (and a
.....Looks like I need a new spoiler (and
.....Looks like I need a new spoiler (and
.....Looks like I need a new spoiler (an
.....Looks like I need a new spoiler (a
.....Looks like I need a new spoiler (
.....Looks like I need a new spoiler
.....Looks like I need a new spoiler
.....Looks like I need a new spoile
.....Looks like I need a new spoil
.....Looks like I need a new spoi
.....Looks like I need a new spo
.....Looks like I need a new sp
.....Looks like I need a new s
.....Looks like I need a new
.....Looks like I need a new
.....Looks like I need a ne
.....Looks like I need a n
.....Looks like I need a
.....Looks like I need a
.....Looks like I need
.....Looks like I need
.....Looks like I nee
.....Looks like I ne
.....Looks like I n
.....Looks like I
.....Looks like I
.....Looks like
.....Looks like
.....Looks lik
.....Looks li
.....Looks l
.....Looks
.....Looks
.....Look
.....Loo
.....Lo
.....L
.....
....
...
..
.

Height = 166667.000000, side = 666666.000000 microns, volume = >74074074073851856.000000 cubic microns.

So 1/6, which is a suspiciously neat fraction.

Suspiciously like a solution d/dx x(1-2x)^2 = 0

-- Richard

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

In article <n7dc1khosulkq2113b729l11n94bodmhls@4ax.com>,
Charlie Roberts <croberts@gmail.com> wrote:

.... Spoiler alert .....

Not sure I get what exactly is the problem here, and being
uncomfortable with numbers, I went about it as follows.

Let the side of the initial sheet be a. Now cut out the sqaures of
side x from the four corners and fold up the sticky bits. You have
an open cistern of side a-2x and height x. The volume is

V(x) = x(a-2x)^2

dV/dx = 0 gives

a - 2x = 0 or a - 6x = 0

The first answer is cute as with x = a/2 leaves with a zero
area base and, naturally, a volume of zero.

The second answer gives x = a/6

Yes

or a volume of a^3/9.

You forgot to square the (a-2x)

-- Richard

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

SPOILER (Tapered case)
SPOILER
SPOILER
SPOILER
SPOILER
SPOILER
SPOILER
SPOILER
SPOILER
SPOILER
SPOILER
SPOILER
SPOILER
SPOILER
SPOILER
SPOILER
SPOILER
SPOILER
SPOILER
SPOILER
SPOILER
SPOILER
SPOILER
SPOILER
SPOILER
SPOILER
SPOILER
SPOILER
SPOILER
SPOILER

y
+-----+----+--
|.....\ |
|......\ |
|-\.....\ |x
y| --\...\ |
| --\.\|
+----------+
| x

If the corners have kite shaped segments removed with the dimensions
shown above,
then the cistern has a bottom side of 1-2*x, a top side of 1-2*x+2*y,
and a height
of sqrt(x^2-y^2).

The volume of a truncated pyramid with bottom side a, top side b and
height h is
V = h*(a^2+a*b+b^2)/3.

With the above dimensions this becomes
V = sqrt(x^2-y^2)*((1-2*x)^2+(1-2*x)*(1-2*x+2*y)+(1-2*x+2*y)^2)/3 =
sqrt(x^2-y^2)*(3-12*x+6*y+12*x^2-12*x*y+4*y^2)/3

For the maximum the partial derivatives with respect to x and y must be
0: x/sqrt(x^2-y^2)*(3-12*x+6*y+12*x^2-12*x*y+4*y^2)+sqrt(x^2-y^2)*(-12+24*x-12*y) = 0 -y/sqrt(x^2-y^2)*(3-12*x+6*y+12*x^2-12*x*y+4*y^2)+sqrt(x^2-y^2)*(6-12*x+8*y)
= 0

Multiplying the first equation by y and the second by x and adding
yields
y*(-12+24*x-12*y)+x*(6-12*x+8*y) = 0
which can be simplified into
3*x-6*y+16*x*y-6*x^2-6*y^2 = 0
and rewritten as
6*x^2-(16*y+3)*x+6*(y+y^2) = 0
This yields
x = (16*y+3-sqrt((16*y+3)^2-144*(y+y^2)))/12 (the + case does not yield
a viable solution)

Multiplying the first equation by x and the second by y and adding
yields
(3-12*x+6*y+12*x^2-12*x*y+4*y^2)+x*(-12+24*x-12*y)+y*(6-12*x+8*y) = 0
which can be simplified into
1-8*x+4*y+12*x^2-12*x*y+4*y^2 = 0
and rewritten as
12*x^2-(8+12*y)*x+(4*y^2+4*y+1) = 0
Substituting x from above yields 12*((16*y+3-sqrt((16*y+3)^2-144*(y+y^2)))/12)^2-(8+12*y)*((16*y+3-sqrt((16*y+3)^2-144*(y+y^2)))/12)+(4*y^2+4*y+1)
= 0
This is 0 for y = 0.123332415765, and then x = 0.232179996793.
This gives a volume of 0.0864197003371.



Please reply to ilanlmayer at gmail dot com

__/\__
\ /
__/\\ //\__ Ilan Mayer
\ /
/__ __\ Toronto, Canada
/__ __\
||

--

--- SoupGate-Win32 v1.05
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On Sun, 4 May 2025 14:15:57 +0000, IlanMayer wrote:

SPOILER (Tapered case)

Amazing. Well done. I need to study that.

I certainly couldn't have done that unasisted, but arrived numerically at
the same volume:

SPOILER (tapered case)
SPOILER
SPOILER
SPOILER
SPOILER
SPOILER
SPOILER
SPOILER
SPOILER
SPOILER
SPOILER
SPOILER
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SPOILER

I can't guarantee this solution, but it looks as though it could be right.

a is the length of the upper edge of the cistern.
x is the length from the corner of the original square sheet to the start
of the cut.
b is the length of the bottom edge of the cistern.
h is the height of the cistern.

The standard square cut-out max-volume solution:

a = 0.667
x = 0.167
b = 0.667
h = 0.167
vol = 0.0741

The tapering fustrum of a pyramid max-volume solution:

a = 0.782
x = 0.109
b = 0.536
h = 0.197
vol = 0.0864

So, about a 17% increase in volume if the sides are allowed to taper.

--
David Entwistle

--- SoupGate-Win32 v1.05
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In article <41858871e4040171463b96c50c238238@www.novabbs.com>,
IlanMayer <ilan_no_spew@hotmail.com> wrote:

[deleted]

Well done!

-- Richard

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On 5/3/2025 1:08 AM, David Entwistle wrote:

Adapted from "Amusements in Mathematics" by Henry Ernest Dudeney.

Given a large sheet of zinc, measuring (before cutting) one metre square,
you are asked to cut out square pieces (all of the same size) from the
four corners of this sheet. You are then to fold up the sides of the resulting shape, solder the edges and make a cistern. The cistern is open
at the top. You can assume you have the appropriate tools and skill to
carry out the task.

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

I guess a follow on question could be; is it possible to cut out four non- square pieces (all of the same size and shape), producing a cistern with tapering sides, which has greater volume. I haven't looked at that yet...

My thoughts about the follow-on puzzle is, "What if the four identical non-square pieces are fractal in nature?". I imagined a first step of
cutting kite-shaped corners using two straight-cuts at each corner
(resulting in a truncated pyramid-shaped cistern). The next refinement
would be to cut curved cuts instead of straight-cuts and bending the
metal in a curve to allow the edges to be soldered together. The
cut-out portion would resemble a kite-shape with two curved sides. The kite-shapes could taper towards the center of the square so that the
volume's cross-section could look like semicircles from two directions
and square from above (but, I'm not sure that this is optimal). Since a portion of the original square is still being cut out and not used,
there may be more room for improvement. What if the curved portions are
not smooth, but zig-zag, creating fine "fingers" along a curve. The
fingers on one side could be soldered to complimentary fingers on the
other side. By adding finer and finer levels of zig-zags to the curve (fractal-like), less and less of the material is wasted. In the limit,
no material would be wasted. The final volume could approach the
maximal volume for the given surface area. The Banach–Tarski paradox
makes me think that one might even be able to get more volume.
--
Carl G.


--
This email has been checked for viruses by AVG antivirus software.
www.avg.com

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

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

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On Wed, 7 May 2025 09:25:15 -0000 (UTC), David Entwistle wrote:

ANSWER

I should have said 'point of inflexion',. not 'inflection', and perhaps
added that h = 1/2 is a local minima, with the resulting volume = 0.

--
David Entwistle

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David Entwistle:

I should have said 'point of inflexion',. not 'inflection'...

Either spelling is fine.

and perhaps added that h = 1/2 is a local minima...

The singular is "minimum".

No big deal, of course.
--
Mark Brader | Those who mourn for "USENET like it was" should
Toronto | remember the original design estimates of maximum msb@vex.net | traffic volume: 2 articles/day. --Steven Bellovin

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