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Local maxima, minima, and saddle points

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    Show that if a>b>c>0 then the function:

     f(x,y,z) = (ax^2 +by^2 + cz^2 )\exp(-x^2-y^2-z^2)

    Can someone please help me with this question I have spent hours on it and have made no progress..
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    Then the function what?
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    Show that if a>b>c>0 then the function:

     f(x,y,z) = (ax^2 +by^2 + cz^2 )\exp(-x^2-y^2-z^2)

    has two local maxima, one local minima and four saddle points.

    Can someone please help me with this question I have spent hours on it and have made no progress..
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    I reckon there is more to it than the second derivative test and algebra. However, I can't think of the right idea yet.
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    What working out have you done so far?
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    (Original post by Neilding91)
    Show that if a>b>c>0 then the function:

     f(x,y,z) = (ax^2 +by^2 + cz^2 )\exp(-x^2-y^2-z^2)

    Can someone please help me with this question I have spent hours on it and have made no progress..
    1. write down the partal derivatives and take them equal zero.
    2. Solve them simoultaneously
    a) first solution is trivial (0,0,0)
    b)then arrange the first equation to x^2 and substitute into the others
    You will get y=0 and z=0 so x^2=1
    this gives 2 soluutions
    c) because of simmetry similary to above for y and z you will get more 4 solutions
    So you have 7 stationary points
    3. Determine the second partial derivatives
    4. Create the Hessian matrix for each given (x,y,z) value (7 matrices)
    5. For every matrix:
    Calculate the eigen values
    a) When the matrix positive definite (all eigen value is positive)
    then there minimum at that point where the matrix was calculated
    b) when the matrix negative definite there is maximum
    c) and when the matrix has both negative and positive eigen values or zero
    then there is a saddle at the given point
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    (Original post by ztibor)
    ...
    That looks like my shopping list -- can't follow it either; too long.
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    I haven't tried it yet but I cant see why brute force wouldn't work in trying to calculate the maxima and minima etc.

    EDIT: as ztbor did. Sounds like a reasonable method.
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    (Original post by gff)
    That looks like my shopping list -- can't follow it either; too long.
    ... can you suggest what to follow
    According the theorem for existence of stationary point in multivariable functions:
    grad( f(\vec r))=\vec 0 (depending on that grad(f) is exists)
    To determine the type of this point you have to use the eigen values for Hessian matrix.
 
 
 
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