Kip in http://www.jsoftware.com/pipermail/programming/2013-February/031712.htmlpresented the argument assuming that the square is perimeter-area optimal for the rectangular polygons. The question is: can one show the implicit assumption without calculus?
I think so: one only has to show that x=c/2 maximizes x(c-x) where the optimal value is (c/2)^2. If one can consider an alternative shifted candidate x= (c/2) + h then x(c-x) = (c/2+h)(c/2 -h) = (c/2)^2 - h^2 <= (c/2)^2 for any h (unless one is considering imaginary shifts). On Thu, Feb 28, 2013 at 12:35 AM, Don & Cathy Kelly <d...@shaw.ca> wrote: > 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 >>>> >>>> ------------------------------**------------------------------** >>>> ---------- >>>> For information about J forums see http://www.jsoftware.com/** >>>> forums.htm <http://www.jsoftware.com/forums.htm> >>>> >>>> >>>> >>>> >>> ------------------------------**------------------------------** >> ---------- >> For information about J forums see >> http://www.jsoftware.com/**forums.htm<http://www.jsoftware.com/forums.htm> >> >> > ------------------------------**------------------------------**---------- > For information about J forums see > http://www.jsoftware.com/**forums.htm<http://www.jsoftware.com/forums.htm> > ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
