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 <rauldmil...@gmail.com> 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 <eel...@pacbell.net> 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 <k...@math.uh.edu> 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 <
>>>> jose.mario.quint...@gmail.com> 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 <k...@math.uh.edu> 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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