r 0 100 _2
50 0
$r 0 100 _2
1 2
Sent from my iPad
On Feb 23, 2013, at 4:35 PM, "Linda Alvord" <[email protected]> wrote:
> The roots of a polynomial:
>
> r=: 13 :'> }. p. y'
> r 0 100 _2 :
> 50 0
> r
> [: > [: }. p.
>
>
> The average of the roots or x coordinate of axis of symmetry:
> a=: 13 :'(+/y)%#y'
> a 50 0
> 25
> a
> +/ % #
>
> Find the maximum:
>
> 0 100 _2 p. 25
> 1250m
>
> Sadly this doesn't work:
>
> a r 0 100 _2
> 50 0
>
> Any idea why not?
>
> Linda
>
>
> -----Original Message-----
> From: [email protected]
> [mailto:[email protected]] On Behalf Of Alex
> Giannakopoulos
> Sent: Saturday, February 23, 2013 3:28 PM
> To: J Programming forum
> Subject: Re: [Jprogramming] The farmer's fence
>
> Nice little gotcha there, assuming that the shape will be a square, since a
> square maximizes the contained area for a rectangle, while forgetting that
> the wall gives you extra perimeter for free, depending on the shape.
>
> By the same analogy I'd tackle Roger's version of the problem, i.e. find
> ANY shape that will maximize the area:
> Again, I suspect that going for a (semi)circle might be essentially the same
> gotcha.
>
> I haven't got time to code it at the moment, but I'd investigate an (half)
> ellipse and also a parabola.
> Will need some integration though, to find the expression for the length of
> their curves.
> ----------------------------------------------------------------------
> For information about J forums see http://www.jsoftware.com/forums.htm
>
> ----------------------------------------------------------------------
> For information about J forums see http://www.jsoftware.com/forums.htm
----------------------------------------------------------------------
For information about J forums see http://www.jsoftware.com/forums.htm
