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Matlab maxima minima problem

  1. Nov 14, 2006 #1
    I have a function that is not giving me critical points for where their appears to be a maximum.

    from plot their seems to be a cp at approximately x=y=5
    [​IMG]

    However, these are the cp's:
    cpx =

    -0.2618
    -1.3090
    1.8326
    2.8798
    -2.8798
    -1.8326
    1.3090
    0.2618


    cpy =

    -0.2618
    -1.3090
    1.8326
    2.8798
    0.2618
    1.3090
    -1.8326
    -2.8798

    When evaluated on the boundry [0,2pi]x[0,2pi] F(0,0)=0 as a absolute minimum and F(2*pi, 2*pi)=12.5664 as absolute maximum...even though in the graph... 2*pi is lower then the imaginary big red mountain on right...for which their seems to be no cp...

    Here's code that generates plot and finds critical points

    Code (Text):

    %plotting it
    [x,y] = meshgrid(0 : .1 : 2*pi)
    z = x+y+4.*sin(x).*sin(y)
    surf(x,y,z)
    %contour(x,y,z,20); axis square; colorbar;

    syms x y
    f = x+y+4*sin(x)*sin(y)
    fx = diff(f,x)
    fxx = diff(fx,x)
    fy = diff(f,y)
    fyy = diff(fy,y)
    fxy = diff(fx,y)
    d = fxx*fyy-fxy^2

    % Solve for all critical points of f using solve
    [cpx,cpy] = solve(fx,fy)

    % Make critical points decimals
    cpx = double(cpx)
    cpy = double(cpy)

    % Make inline functions for f, fxx, and d
    F = inline(vectorize(f),'x','y')
    D = inline(vectorize(d),'x','y')
    Fxx = inline(vectorize(fxx),'x','y')

    % define boundries
    %[0 0; 0 2*pi; 2*pi 0; 2*pi 2*pi]
    boundx = [0; 0; 2*pi; 2*pi]
    boundy = [0; 2*pi; 0; 2*pi]

    % Make a table of the cp's, F(at cp's), D(at cp's), and Fxx(at cp's)
    T = [cpx cpy F(cpx,cpy) D(cpx,cpy) Fxx(cpx,cpy)]

    %Evaluate F at boundries of region [0,2pi]x[0,2pi]
    T = [boundx boundy F(boundx,boundy)]
     
    So my question is....is their a cp at that red mountain? or is their no red mountain :P
    They say when fx and fy DNE their is also cp...but i don't see that happening since fx and fy are polynomials...
     
    Last edited: Nov 14, 2006
  2. jcsd
  3. Nov 14, 2006 #2

    JasonRox

    User Avatar
    Homework Helper
    Gold Member

    Are you solving for fx = fy? Or better yet fx - fy = 0?

    I kind of see that in your little problem, but everything after that seems like a blur.

    Why don't you just solve the above expression, and check if those critical points are maximums?

    I'm not sure if that is exactly what the graph looks like, so I'll assume you did it right. If you did it right, it seems like you do have a maximum. Now the problem is either the program can't solve it or the steps you're following are wrong.
     
  4. Nov 14, 2006 #3
    Yea it looks like it's solving them equal to each other or considering both...same thing. It's how an example i found did it for these sorts of problems.

    Mathematica gave same answers for critical points.

    Solve[{0.0 == Fx[x, y], 0.0 == Fy[x, y]}, {x, y}]

    {{y -> -2.87979, x -> 0.261799}, {
    y -> -1.8326, x -> 1.309}, {y -> -1.309, x -> -1.309}, {y -> -0.261799,
    x -> -0.261799}, {y -> 0.261799, x -> -2.87979}, {y ->
    1.309, x -> -1.8326}, {y -> 1.8326, x -> 1.8326}, {
    y -> 2.87979, x -> 2.87979}}
     
  5. Nov 14, 2006 #4
    Hmm i think i got em...was reading some warning messages in mathematica that saying not all solutions may be found cause it was using inverse functions, and to use some function called Reduce...so with a bit of tinkering...

    N[Reduce[{0 == Fx[x, y] , 0 == Fy[x, y], 0 ≤ x ≤ 2*\[Pi] , 0 ≤ y ≤ 2*\[Pi]}, {x, y}]]

    finds these:

    (x == 4.97419 && y == 4.97419) || (x == 6.02139 && y == 6.02139) || (x ==
    1.8326 && y == 1.8326) || (
    x == 2.87979 && y == 2.87979) || (x == 3.40339 && y == 0.261799) || (x == 4.45059 && y == 1.309) || (x == 0.261799 && y == 3.40339) || (x ==
    1.309 && y == 4.45059)

    and those look like the cp's i was looking for.
     
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