Among the many answers in this thread 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 <[email protected]> >Til: [email protected] >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 > >---------------------------------------------------------------------- >For information about J forums see http://www.jsoftware.com/forums.htm > > > ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
