Hi J'ers
The communication below was sent to, but seemingly not received by, 
<programm...@jsoftware.com>.
So I resend itfor your information.
- Bo





>________________________________
> Fra: Bo Jacoby <bojac...@yahoo.dk>
>Til: "programm...@jsoftware.com" <programm...@jsoftware.com> 
>Sendt: 20:05 søndag den 24. februar 2013
>Emne: SV: [Jprogramming] The farmer's fence
> 
>
>Among the many answers in this thread about the farmer's fence the standard 
>method is not seen, so here it comes.
>
>
>Let the sides of the rectangular chicken yard be  x  and  y
>The area is  x*yDifferentiating the area gives  0=(y*dx)+(x*dy)  which is zero 
>because the area is maximum.
>The length of fence is  100= (2*x)+y
>Differentiating the length gives  0=(2*dx)+(dy)  which is zero because the 
>lenght is constant. 
>
>Multiplying   0=(2*dx)+(dy)  by  x  (called a Lagrange multiplyer) gives  
>0=(2*x*dx)+(x*dy)
>
>Subtracting   0=(2*x*dx)+(x*dy)   from   0=(y*dx)+(x*dy)  gives   0=(y-2*x)*dx
>As  dx  need not be zero we must have  0=y-2*x .  
>
>The rest is easy.
>
>So  y=2*x . 
>
>Substitute  y=2*x   into 100=(2*x)+y  and get  100=4*x. 
>
>Divide by  4  and get   x=25 
>
>Substitute  x=25  into  100=(2*x)+y  and  get  y=50. 
>
>
>
>No computer power is required.
>- Bo
>
>
>>________________________________
>> Fra: km <k...@math.uh.edu>
>>Til: programm...@jsoftware.com 
>>Sendt: 15:42 lørdag den 23. februar 2013
>>Emne: [Jprogramming] The farmer's fence
>> 
>>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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>>
>>
>>
>
>
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