If we let the machine do the job I can offer a specialized verb:
area=: [: (p. 1&{::@p.@p..) 0 1 +//.@(*/) _1r2,~-:
... and here's a verb that calculates the area of a circle segment
depending on the cut off part off the circle
SegmentArea=: ([: (-1&o.) 2p1%1+%) (1p1&*@*:@]-(*-:@*:)) 2p1 %~ +
use:
(wire fence length) SegmentArea arc length cut off segment
SegmentArea~ 100
1591.55
>./ 100 SegmentArea 90+0.01*i.2000
1591.55
plot (;100&SegmentArea) 90+0.01*i.2000
On 23-02-13 19:46, km 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
On Feb 23, 2013, at 12:03 PM, Aai <agroeneveld...@gmail.com> wrote:
With some math in J
x : length barn side
-->the other rectangle side is then
-: 100 - x
i.o.w. we have
A = x (50 - x/2)
with J
0 1 +//.@(*/) 50 _0.5
0 50 _0.5
first derivative
0 50 _0.5&p. d. 1
50 _1&p.
second derivative
0 50 _0.5&p. d. 2
_1"0
--> a maximum for
1{:: p. 50 _1
50
--> maximum (rectangular) area
0 50 _0.5&p. 50
1250
On 23-02-13 15:42, km wrote:
Use J to solve the farmer's fence problem:
A farmer with 100 meters of wire fence wants to make a rectangular chicken yard
using an existing barn wall for one of the north-south sides. What is the
largest area he can enclose if he uses the 100 meters of fence for the other
three sides, and what are the dimensions of the largest-area chicken yard?
Kip Murray
Sent from my iPad
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