try
r 100 _4
25
You want the root of the derivative of p.0 100 _2 to be 0 -that is p.100 _4
0 100 _2 p. r 100 _4
1250
Don
On 23/02/2013 2:35 PM, Linda Alvord 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: programming-boun...@forums.jsoftware.com
[mailto:programming-boun...@forums.jsoftware.com] 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.
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