I believe you are right. I waver back and forth as to whether or not
your explanation constitutes a proof. I know(?) that therectangular one
is correct because it is a simple min-max problem (set the first
derivative of the ratio of area to perimeter to zero). When I did this,
I got your result. Thank you; you set me to thinking.
P.S. While typing this, I just got a wild idea about a possible proof
notusing calculus of variations. If it comes to anything, I will let you
know. If not, I am through with farmer's fences.
On 02/24/2013 07:08 AM, km wrote:
Without getting into calculus of variations, we can think of combining the
rectangular chicken yard with its mirror image on the other side of the barn
wall. The result is a rectangle of perimeter 200. We saw that the
corresponding largest area rectangle was a 50 by 50 square. So perhaps the the
largest area chicken yard using the barn wall and 100 meters of fence is a
semi-circle (combining with its mirror image to make a circle of perimeter
200). Taking for granted that a circle is the figure of largest area having a
given perimeter.
Sent from my iPad
On Feb 23, 2013, at 9:56 PM, Raul Miller <[email protected]> wrote:
I was thinking that raindrop formation on windows would suggest some
shape that is approximately circular.
If we have a half circle, the circle's radius should be
R=: 100%1p1
and the half circle's area should be:
(o.R^2)%2
1591.55
So if there's a better shape it must have a larger area.
I expect that a shape with larger area would have a maximum distance
from the wall which is shorter than R but I would have to think a bit
about how to characterize where the resulting extra length belongs.
That said... I've not worked with calculus of variations and the
wikipedia article does not come with enough concrete examples for me
to figure out what it is that I do not understand.
FYI,
--
Raul
On Sat, Feb 23, 2013 at 10:24 PM, Eldon Eller <[email protected]> wrote:
Old and senile as I am, this looks to me like a problem in calculus of
variations.See, e.g., en. wikipedia.org/wiki/Calculus_of_variations. You are
not likely to get the solution by guessing that the shape is elliptical, or
catenary, or parabolic, etc. I am too old and lazy to try to solve it
myself. I would like to see someof you who are smarter and more energetic
than I give it a go. I feel reasonably certain that the problem has a closed
form solution and that writing that out in J would not be difficult. What
would be reallyimpressive would be a numerical method of doing calculus of
variations, in J, of course.
On 2/23/2013 4:28 PM, Ric Sherlock wrote:
Kip,
Alternative formulations for your adverb that require fewer calculations
of
u y.
Max1 =: 1 : 0
((= >./)@:u # ] ,. u) y
)
Max2 =: 1 : 0
fnres=. u y
where=. (= >./) fnres
where # y ,. fnres
)
I'd be interested in a tacit implementation of one of the adverbs Max
above. I came up with the same as Pepe for a simple Max ( >./@: ) but
can't see how to "factor out" the verb area from the adverbs in the
following:
Max3=: (({~ area ((i. >./)@:)) , area (>./@:))
On Sun, Feb 24, 2013 at 9:18 AM, km <[email protected]> wrote:
Borrowing ideas from Raul, I like
Max =: 1 : 0
max =. >./ u y
where =. max = u y
where # y ,. u y
)
which identifies the max and where it occurs:
*: Max i:2
_2 4
2 4
(4 - *:) Max i:2
0 4
Sent from my iPad
On Feb 23, 2013, at 1:04 PM, Jose Mario Quintana <
[email protected]> wrote:
I did not see your second post!
area=. ] * 50 - %&2
area(max=. (>./) @:) 0 to 100
1250
max
./@:
On Sat, Feb 23, 2013 at 1:46 PM, km <[email protected]> wrote:
Can we have an adverb Max so that f Max y finds the maximum of f on
the list y ?
Sent from my iPad
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