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Must find a function with specified max and min

  1. Oct 21, 2007 #1
    This is a calc 3 problem dealing with partial derivatives in 3space:

    I know that for a function to have a max at a specified point (a,b):

    Fx(a,b)=0
    Fy(a,b)=0
    D>0 where D=Fxx(a,b)Fyy(a,b)-[Fxy(a,b)]^2
    D can not equal 0
    Fxx(a,b)<0

    For a min at specified point:

    Fx(a,b)=0
    Fy(a,b)=0
    D>0 where D=Fxx(a,b)Fyy(a,b)-[Fxy(a,b)]^2
    D can not equal 0
    Fxx(a,b)>0


    So i've been spending the past few hours trying to figure out this problem and i'm not exactly sure how to go about doing it. I've just been making up functions by trial and error and havent been getting anywhere really.

    The one function that comes closest that I came up with was:

    (x^2)(y^3)-(x^4)(y^6)+1/2(x^4)(y^6)

    since x and y partials are both equal to 0 at point (1,1) and Fxx partial is negative... but when i plug in for D I get 0.

    So what is the proper way of approaching this problem?
     
    Last edited: Oct 21, 2007
  2. jcsd
  3. Oct 21, 2007 #2
    My guess at a simple function would be

    y=a*[(x-b)^3-(x-b)^2]+c

    with a, b c to be determined by differentiation.
     
  4. Oct 21, 2007 #3
    I am a bit confused with how you came up with that equation. I need a function of two variables (x,y)... I think?

    Something similar to this : f(x,y)=(x^2)(y^3)-(x^4)(y^6)+1/2(x^4)(y^6)

    Thanks for your help.
     
    Last edited: Oct 21, 2007
  5. Oct 21, 2007 #4
    A function of two variables defines a surface z=f(x,y). For every point in the x-y plane the function value gives its 'height'.
     
  6. Oct 21, 2007 #5

    arildno

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    Well, it is a two-variable function you're after.

    So, because you know that the second derivatives of quadratic functions equal some constant, they are clearly unsuitable, because the Hessian is needed to change its value at times.

    Try therefore for a general cubic:
    [tex]f(x,y)=ax^{3}+bx^{2}y+cxy^{2}+dy^{3}+ex^{2}+fxy+gy^{2}[/tex]
    Since you are essentially dealing with six restrictions, you should be able to determine paremeters a,b,c,d, e,f,g wirhout having to take into account linear terms as well.
     
    Last edited: Oct 21, 2007
  7. Oct 21, 2007 #6

    arildno

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    Solve the four linear equations first, getting two free variables.

    Then fiddle with those variables so that the Hessians at both points have the correct signs.
     
  8. Oct 21, 2007 #7
    OK- it sounds like I'm confused. A max at (1,1) could mean a max at y(1)=1, or it could mean a max at f(1,1).
     
  9. Oct 21, 2007 #8

    arildno

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    The four equations are:
    [tex]3a+2b+c+2e+f=0[/tex] (fx at (1,1))
    [tex]b+2c+3d+f+2g=0[/tex] (fy at (1,1))
    [tex]3a+2b+c-2e-f=0[/tex] (fx at (-1,-1))
    [tex]b+2c+3d-f-2g=0[/tex] (fy at (-1,-1))

    Probably, you can set e=f=g=0; and solve the resultantant system:
    3a+2b+c=0
    b+2c+3d=0

    whereby c=-3a-2b, yielding d=2a+b
     
    Last edited: Oct 21, 2007
  10. Oct 21, 2007 #9
    Sorry for not being clear. I need a max at f(1,1) and a min at f(-1,-1).
     
  11. Oct 21, 2007 #10
    Ok so the function has to be a cubic? Forgive me for being slow, I'm having a hard time understanding whats going on lol.
     
  12. Oct 21, 2007 #11

    arildno

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    Of course it doesn't have to be cubic!

    There are zillions of functions satisfying your criteria.

    However, out of CONVENIENCE, instead of looking after exotic functions that might satisfy your criteria, I choose to look for a boring polynomial because it is easy to deal with.

    Understand the difference?
     
  13. Oct 21, 2007 #12
    Yes I understand. After I solve for all the coefficients a,b,c,d,e,f,g(still working on it) isnt there other criteria that must be met? Since you just set all the partials to 0 at the points of interest that just tells you they are critical points. From there they could be max's, mins or saddle points. Right? I'm still working it out now, but thank you for your help.
     
  14. Oct 21, 2007 #13

    arildno

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    Correct!
    Right!
    But tackle the inequalities at the end.
    I am quite sure you won't need e,f,g to be any other numbers than zero.
    The reason for this is that even without them, you'll get two free parameters out of the partials equations. That should give you enough fiddling space to satisfy the Hessian demands.

    Good luck to you!



    The rest of your criteria is that your function should be smooth&continuous, which is automatically satisfied by all polynomials.
     
  15. Oct 21, 2007 #14
    I can't figure out the coefficients. :( I know this is just algebra and I should know it, but how do you solve for the 4 unknowns(a,b,c,d) with only the 2 equations(Fx,Fy)?
     
  16. Oct 21, 2007 #15
    I think i'm heading down the right track with the Cubic function thanks to arildno but I can't figure out the coefficients. Anyone else out there that could help?
     
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