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Math Help - Help with Stationary Points of a surface sadle points thing ..

  1. #1
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    Help with Stationary Points of a surface sadle points thing ..

    i have an exam coming up and this is going to be in it i have a problem in my notes i cant seem to do can anyone help please

    Find the stationary points of the following surface. and determine thier nature of the stationary points.

    f(x,y) = x^3 + y^3 - 3x - 12y +20

    i understand the steps but for some reason when a Partial diff it never seems to come out like anything els in my notes any help would be very much apreciated
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    Quote Originally Posted by AVVIT
    i have an exam coming up and this is going to be in it i have a problem in my notes i cant seem to do can anyone help please

    Find the stationary points of the following surface. and determine thier nature of the stationary points.

    f(x,y) = x^3 + y^3 - 3x - 12y +20

    i understand the steps but for some reason when a Partial diff it never seems to come out like anything els in my notes any help would be very much apreciated
    I might be totally wrong but it seems to me to find the critical point for this function.

    Thus, you need that,
    \left\{ \begin{array}{c}\frac{\partial f}{\partial x}=0\\ \, \\\frac{\partial f}{\partial y}=0
    But,
    \frac{\partial f}{\partial x}=3x^2-3
    \frac{\partial f}{\partial y}=3y^2-12

    Solving the equations we have,
    x=-1,1
    y=-2,2
    Thus, the only possible points are,
    (-1,-2,38),(-1,2,-6),(1,-2,34),(1,2,2)
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  3. #3
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    This is the only problem thats one of the answers i get but it cant be write as its suposed to be harder than that lol... or so i thought examples in my notes when partial differentiated always end up with an x and Y in the resultant making them harder and then transposing the forumula to calc maximas and minimas ..
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    To determine whether these points are maximum or minimum you need to find the value of,
    D=\left| \begin{array}{cc} f_{xx}& f_{xy}\\ f_{xy}& f_{yy} \end{array} \right|

    We find that,
    f_{xx}=6x^2
    f_{yy}=6y^2
    f_{xy}=0
    Therefore,
    D=\left| \begin{array}{cc} 6x^2&0\\ 0&6y^2 \end{array} \right|=36x^2y^2>0
    Looking at,
    f_{xx}(x_0,y_0)>0
    Thus it is a relative minimuma.
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