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Math Help - Local minimum and maximum and saddle points problem

  1. #1
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    Exclamation Local minimum and maximum and saddle points problem

    I'm wondering that I did this problem correct, or not.

    Find the local maximum and minimum values and saddle point(s) of the function f(x,y) = 1+2xy-x^2-y^2
    \nabla f = 0
    \nabla f =<\frac{\partial f }{\partial x },\frac{\partial f }{\partial y }>=<2y-2x,2x-2y>

    Then, I got x=y. After that, I tried to find D = fxx*fyy-f^2xy
    I've got 0 from D. Therefore, I concluded D=0, no conclusion.

    Am I right? Is there anything wrong?

    Thank you.
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  2. #2
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    Quote Originally Posted by noppawit View Post
    I'm wondering that I did this problem correct, or not.

    Find the local maximum and minimum values and saddle point(s) of the function [tex]f(x,y) = 1-x^2-y^2[/+2xymath]
    \nabla f = 0
    \nabla f =<\frac{\partial f }{\partial x },\frac{\partial f }{\partial y }>=<2y-2x,2x-2y>

    Then, I got x=y. After that, I tried to find D = fxx*fyy-f^2xy
    I've got 0 from D. Therefore, I concluded D=0, no conclusion.

    Am I right? Is there anything wrong?

    Thank you.
    If  D = 0 , then examine the function:

    \Delta(h,k) = f(a+h,b+k)-f(a,b)

    If \Delta(h,k) > 0 \forall h,k , then you have a local minimum.

    If \Delta(h,k) < 0 \forall h,k , then you have a local maximum.

    If \Delta(h,k) < 0 for SOME h,k, and \Delta(h,k) > 0  , for OTHER h,k then you have a saddle point.

     \Delta(h,k) = 1+2(a+h)(b+k) -(a+h)^2-(b+k)^2 - (1-a^2-b^2)

     \Delta(h,k) = 1+2(ab+ak+hb+hk)-(a^2 + 2ah +h^2)-(b^2+2bk+k^2) - 1+a^2+b^2

     \Delta(h,k) = 2ab+2ak+2hb+2hk - 2ah -h^2-2bk-k^2

     \Delta(h,k) = 2(ab+ak+ah+hk+hb-bk-ah)-h^2-k^2

     \Delta(h,k) =  2(ab+ak+hk+hb-bk)-h^2-k^2

    You must plug in a and b. These are the points where you find stationary points to be. After that you should be able to conclude that:

     \Delta(h,k) is > 0 for some h k, and < 0 for other. This indicates stationary points.
    Last edited by Mush; January 28th 2009 at 03:18 AM.
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  3. #3
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    Quote Originally Posted by noppawit View Post
    I'm wondering that I did this problem correct, or not.

    Find the local maximum and minimum values and saddle point(s) of the function f(x,y) = 1+2xy-x^2-y^2
    \nabla f = 0
    \nabla f =<\frac{\partial f }{\partial x },\frac{\partial f }{\partial y }>=<2y-2x,2x-2y>

    Then, I got x=y. After that, I tried to find D = fxx*fyy-f^2xy
    I've got 0 from D. Therefore, I concluded D=0, no conclusion.

    Am I right? Is there anything wrong?

    Thank you.
    You'll notice that your function can be written as

    f(x,y) = 1 - (x-y)^2

    and since the second term is always negative then

    f(x,y) = 1 - (x-y)^2 \le 1

    so the max if f is located on the line  y = x and is 1.
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  4. #4
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    Hello, noppawit!

    I'd say your work and conclusion are correct.

    Note that if x = y, the function is: . f(x,x) \:=\:1 + 2x^2 - x^2 - x^2 \:=\:1

    The function is constant above the line y = x.

    So there is a horizontal tangent plane: z = 1

    Evidently none of those points (x,x,1) is a maximum/minimum/saddle point.

    (Well, they could be ... but I'm not sure.)
    .
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