I did this a bit more simply- before seeing your version:
100=2*x+y or y=100-2*x (treating y as dependent on x)
A=xy=100x-2*x^2
or dA/dx= 100-4x =0 using introductory calculus.
x=25
I very much like the alternative that was given by someone else- noting
that the square has the largest area of a quadrilateral consider a
square with a perimeter of 200m - or 50mby 50m then take half of it. No
calculus needed-(except by the guy who figured out the square (or
rhombus) is the quadrilateral with the largest area) -just thinking.
cheers
Don
On 27/02/2013 3:59 PM, Bo Jacoby wrote:
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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