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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