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Homework Help: Maxima/Minima of extrema function f(x,y)=x

  1. Mar 22, 2010 #1
    1. The problem statement, all variables and given/known data
    Find the extrema of the function subject to the given constraint.

    f (x, y) = x

    x2 + 2y2 = 3


    2. Relevant equations
    det = fxx * fyy - (fxy)2
    If det > 0,
    fxx < 0 [tex]\Rightarrow[/tex] MAXIMUM
    fxx > 0 [tex]\Rightarrow[/tex] MINIMUM

    If det < 0,
    [tex]\Rightarrow[/tex] SADDLE POINT

    If det = 0m
    Inconclusive

    3. The attempt at a solution

    I appear to be struggling with this maxima/minima stuff. Previous questions were just correct. I had to look at the book's answers to figure a way there. I seem to be missing something...but here is my attempt:

    Since x2 + 2y2 = 3,

    x = [tex]\pm[/tex][tex]\sqrt{3 - 2y2}[/tex]

    Therefore,

    f(x, y) = [tex]\pm[/tex][tex]\sqrt{3 - 2y2}[/tex]

    At critical point,

    [tex]\nabla[/tex] f = (0, - [tex]\frac{2y}{\sqrt{3 - 2y2}}[/tex] )

    = (0, 0)


    Therefore, equating gives

    0 = 0

    &

    - [tex]\frac{2y}{\sqrt{3 - 2y2}}[/tex] = 0

    So extrema points at ([tex]\sqrt{3}[/tex], 0) & (-[tex]\sqrt{3}[/tex], 0) since y = 0.


    To find max/min/saddle, find det.

    fxx = 0
    fxy = 0

    det = fxx * fyy - (fxy)2
    det = 0 * fyy - (0)2
    det = 0


    Which is inconclusive. The book has ([tex]\sqrt{3}[/tex], 0) as a maximum and (-[tex]\sqrt{3}[/tex], 0) as a minimum.

    I don't know what I'm missing out here...

    Thanks in advance.
     
  2. jcsd
  3. Mar 22, 2010 #2

    Mark44

    Staff: Mentor

    Tip: Don't mix [ sup] or [ sub] tags inside [ tex] tags. Instead use ^{} for exponents and _{} for subscripts.
    I don't see how the equation above is helpful at all.
    What I think you are missing is some understanding of what f(x, y) = x looks like, which is a plane. Since a plane is a flat surface with no curvature, there are no dips or bumps. Also, since the domain of this plane is restricted to {(x, y) | x2 + 2y2 = 3}, any maximum or minimum points are attained at points the farthest away from the origin.
     
  4. Mar 22, 2010 #3
    Ohh...I see... I feel so silly ><

    Thanks a lot for that.



    If I may, I have another question regarding maxima and minima...

    The problem statement, all variables and given/known data

    Find local minima, maxima and saddle points of the function:

    f(x, y) = (x2 + 3y2) e1-x2-y2


    I started off finding the partial derivatives of x and y respectively and got:

    fx = 2xe1-x2-y2 ( 1 - x2 - 3y2 )
    fy = 2ye1-x2-y2 (3 - x2 - 3y2 )


    At critical points, (fx, fy) = (0, 0)

    Equating gives,

    2xe1-x2-y2 ( 1 - x2 - 3y2 ) = 0 --- (1)
    2ye1-x2-y2 (3 - x2 - 3y2 )= 0 --- (2)



    After that I'm kinda lost...I slithered my way to get (0, 0) but very unconvincingly. I would rather have some solid reasoning as to why.

    Like I mentioned earlier, I think I'm missing some understanding and maybe that's why I'm not getting this.

    The answers give
    Min at (0, 0),
    max at (0, 1), (0, -1),
    saddle at (-1, 0), (1, 0)

    Thanks!
     
  5. Mar 22, 2010 #4

    vela

    User Avatar
    Staff Emeritus
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    When you have a product equal to 0, then one of the factors must be equal to 0. For instance, from this equation

    [tex]2xe^{1-x^2-y^2}(1-x^2-3y^2) = 0[/tex]

    you can conclude that x=0 or 1-x2-3y2=0. Similarly, the other equation gives you y=0 or 3-x2-3y2=0. Now you just have to find the solutions for all four possible combinations:

    (1) x=0, y=0;
    (2) x=0, 3-x2-3y2=0;
    (3) y=0, 1-x2-3y2=0; and
    (4) 1-x2-3y2=0, 3-x2-3y2=0.
     
  6. Mar 22, 2010 #5
    Ooo, I see now. I think I managed to confuse myself by using how for eq (1),

    since x = 0,

    1-x2-3y2=0
    1 - 3y2 = 0

    which is definitely not looking good in terms of the answers. But I understand now =)

    One last thing, so for the 4th solution: (4) 1-x2-3y2=0, 3-x2-3y2=0.

    There isn't a solution for this yeah? Since all the variables x/y cancel out.
     
  7. Mar 22, 2010 #6

    vela

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    Right idea, but wrong equation. That's the one that has to hold if y is 0 and x isn't 0.
    Yup, that combination has no solutions.
     
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