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Math Help - local max, min and saddle points

  1. #1
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    local max, min and saddle points

    ok, hello!

    Determine all local minimum, maximum and saddle points for f(x,y)=x^3-y^3-15xy.
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  2. #2
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    Quote Originally Posted by jacek View Post
    ok, hello!

    Determine all local minimum, maximum and saddle points for f(x,y)=x^3-y^3-15xy.
    This is not in the form y=f(x)= function of x.
    y is part of the equation itself.

    What value is this equal to ?
    Is it zero? or some constant ?
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  3. #3
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    That's whole content of the task.
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  4. #4
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    Sorry,
    it's a three dimensional function,
    not y=function of x,
    I haven't done these before...
    Last edited by Archie Meade; February 9th 2010 at 06:18 AM.
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  5. #5
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    This is not a question of "find the max and min of y as a function of x", it is "find the max and min of f(x,y)". That is, think of z= f(x,y) and find critical points of z.

    The critical points of a function of two variables are where \nabla f= 0 which is the same as saying [tex]\frac{\partial f}{\partial x}= 0[tex] and \frac{\partial f}{\partial y}= 0.

    The "second derivative test" for a function of two variables is more complicated than for a function of one variable. First calculate
    \left|\begin{array}{cc}\frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x\partial y} \\ \frac{\partial^2 f}{\partial y\partial x} & \frac{\partial^2 f}{\partial y^2}\end{array}\right|= \left(\frac{\partial^2f}{\partial x^2}\right)\left(\frac{\partial^2 f}{\partial y^2}\right)- \left(\frac{\partial^2 f}{\partial x\partial y}\right)^2.

    If D> 0, then the point is a maximum if \frac{\partial^2 f}{\partial x^2} is positive and a minimum if it is negative. If D< 0, then the point is a saddle point.
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  6. #6
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    Ah yes! Thanks HallsofIvy,

    Z=f(x,y)=x^3-y^3-15xy

    A critical point occurs when both partial derivatives are zero.

    \frac{\partial f(x,y)}{\partial x}=3x^2-15y=0\ for\ x^2=5y\ \Rightarrow\ y=\frac{x^2}{5}

    \frac{\partial f(x,y)}{\partial y}=-3y^2-15x=0\ for\ y^2=-5x\ \Rightarrow\ x=-\frac{y^2}{5}

    Hence, we solve the resulting pair of equations for x and y.

    y=\frac{y^4}{125}\ \Rightarrow\ 125y=y^4\ \Rightarrow\ y\left(y^3-125\right)=0\ \Rightarrow\ y=0\ and\ 5

    y=0\ \Rightarrow\ x=0
    y=5\ \Rightarrow\ x=-5

    The critical points are (0,0) and (-5,5).

    To determine the nature of the critical points, the second partial derivatives are examined.

    Also required is \frac{\partial}{\partial y}\ \frac{\partial f(x,y)}{\partial x}

    \frac{\partial^2 f(x,y)}{\partial x^2}=6x

    \frac{\partial^2 f(x,y)}{\partial y^2}=-6y

    \frac{\partial}{\partial y}\ \frac{\partial f(x,y)}{\partial x}=-15


    D=\frac{\partial^2 f(x,y)}{\partial x^2}\ \frac{\partial^2 f(x,y)}{\partial y^2}-\left(\frac{\partial}{\partial y}\ \frac{\partial f(x,y)}{\partial x}\right)^2=-36xy-(-15)^2=-36xy-225

    D>0 and \ \color{blue}\frac{\partial^2 f(x,y)}{\partial x^2}>0\ \Rightarrow\ relative minimum point

    D>0 and \ \color{blue}\frac{\partial^2 f(x,y)}{\partial x^2}<0\ \Rightarrow\ relative maximum point

    D<0 \color{blue}\Rightarrow\ saddle point

    At (0,0) D<0 therefore this is a saddle point.

    At (-5,5) D is 900-225 >0 and \frac{\partial^2 f(x,y)}{\partial x^2} = -30 <0 so this is a relative maximum.
    Attached Thumbnails Attached Thumbnails local max, min and saddle points-graph-3d.jpg   local max, min and saddle points-graph-3d2.jpg  
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  7. #7
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    on the very end, why do we use only second derivative of x?
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  8. #8
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    The 2nd derivative of y is also utilised,
    the results for "D", the "discriminant" are determined by the equation for D,
    which is regarded as a theorem.

    The equation for D does involve the product of both 2nd derivatives.

    The saddle points have a "local maximum for x" and a "local minimum for y" colliding at a point or vice versa, which is a way of putting it.
    Hence no maximum for both x and y simultaneously (or maximum for both) to identify the point as either a maximum or minimum.

    D<0 for a saddle point.
    If D>0, then a critical point will be a maximum or minimum.

    The reason is based on the formula for D.

    D=\frac{\partial^2 f}{\partial x^2}\frac{\partial^2 f}{\partial y^2}-\left(\frac{\partial^2 f}{\partial x\partial y}\right)^2

    Since the RHS of the above expression is always positive,
    then the only way D can be >0 is for both 2nd derivatives to be simultaneously positive or simultaneously negative.
    Hence, if D>0, there is no need to double-check both 2nd derivatives.
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  9. #9
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    Can anyone check if this is correct?

    f(x,y)=6x^2y-12y^2-x^4+18y

    two points:
    x=3 y=3
    x=-3 y=3

    Discriminant is: -y-5x^2
    And for both points we get -48, so there's only one saddle point and no local max. or min.

    Please check it!
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  10. #10
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    Quote Originally Posted by jacek View Post
    Can anyone check if this is correct?

    f(x,y)=6x^2y-12y^2-x^4+18y

    two points:
    x=3 y=3
    x=-3 y=3

    Discriminant is: -y-5x^2
    And for both points we get -48, so there's only one saddle point and no local max. or min.

    Please check it!
    hi jacek,

    there are 3 critical points.

    \frac{\partial}{\partial_x}f(x,y)=12xy-4x^3=0\ \Rightarrow\ 3xy-x^3=0\ \Rightarrow\ x\left(3y-x^2\right)=0

    x=0,\ y=\frac{x^2}{3}

    \frac{\partial}{\partial_y}f(x,y)=6x^2-24y+18=0\ \Rightarrow\ x^2+3=4y\ \Rightarrow\ y=\frac{x^2+3}{4}

    This yields the 3 critical points (0,0), (-3,3), (3,3).

    \frac{\partial^2}{\partial_x}f(x,y)=12y-12x^2

    \frac{\partial^2}{\partial_y}f(x,y)=-24

    \frac{\partial^2}{\partial_x\partial_y}f(x,y)=12x

    D=12\left(y-x^2\right)(-24)-(12x)^2

    At (0,0), D=0, cannot be identified as a saddle point using the 2nd derivative test.

    At (-3,3)\ D>0,\ f_{xx}<0 local maximum

    At (3,3)\  D>0,\ f_{xx}<0 local maximum
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  11. #11
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    So, when it comes to discriminant, I cannot cancel 12's from 12y-12x^2 to make it y-x^2?
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  12. #12
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    Quote Originally Posted by jacek View Post
    So, when it comes to discriminant, I cannot cancel 12's from 12y-12x^2 to make it y-x^2?
    No, you can't.
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