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Math Help - Critical points for multivariable function

  1. #1
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    Critical points for multivariable function

    The last part of this problem was confusing me, so any help would be appreciated:

    Here's the problem:

    For the surface defined by the function F(x,y) = 2(x^2)y - y^2 - 4x^2 + 3y, find and classify all critical points.

    So far, I've taken the partial derivatives with respect to x and y and have gotten

    4xy-8x for the derivative with respect to x and
    2x^2 - 2y + 3 for the derivative with respect to y

    From those two equations, I got my critical points (by setting the equations = to 0)

    For my critical points, I have (+ the square root of .5, 2) and (- the square root of .5, 2).

    What I'm confused about is how to classify the points (meaning whether they are minimum, maximum, saddle points, etc.).
    Last edited by clockingly; March 28th 2007 at 01:19 PM.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by clockingly View Post
    This problem was confusing me so any help would be appreciated:

    Here's the problem:

    For the surface defined by the function F(x,y) = 2(x^2)y - y^2 - 4x^2 + 3y, find and classify all critical points.

    So far, I've taken the partial derivatives with respect to x and y and have gotten

    4xy-8x for the derivative with respect to x and
    2x^2 - 2y + 3 for the derivative with respect to y

    I know that I have to set these two equations equal to 0, but I'm confused as to what exactly the critical points are.

    Meaning are they the (x,y) from the first equation and the (x,y) from the second equation?

    I've solved for y from that first equation and got that it = 2.
    I've also solved for x for that second equation and got that it is equal to +-the square root of .5
    critical points in multivariable function are either maximums, minimums or saddle points.

    here is the method:

    given f(x,y),
    find fx, fy, fxx, fyy,fxy

    to find the critical points, set fx = 0, and fy = 0
    doing this you will obtain one or more points, say (x0, y0), (x1,y1),...

    then set up the discriminant equation:

    D(x0,y0) = fxx (x0,y0) * fyy(x0,y0) - (fxy(x0,y0))^2

    if D>0, and fxx(x0,y0)> 0 we have a local min

    if D>0, and fxx(x0,y0)< 0 we have a local max

    if D< 0 we have a saddle point
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    critical points in multivariable function are either maximums, minimums or saddle points.

    here is the method:

    given f(x,y),
    find fx, fy, fxx, fyy,fxy

    to find the critical points, set fx = 0, and fy = 0
    doing this you will obtain one or more points, say (x0, y0), (x1,y1),...

    then set up the discriminant equation:

    D(x0,y0) = fxx (x0,y0) * fyy(x0,y0) - (fxy(x0,y0))^2

    if D>0, and fxx(x0,y0)> 0 we have a local min

    if D>0, and fxx(x0,y0)< 0 we have a local max

    if D< 0 we have a saddle point
    I have edited my first post to include the critical points that I got. Here they are below:

    I have (+ the square root of .5, 2) and (- the square root of .5, 2).

    Would this mean that the first point is a max and the second point is a min?
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