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Absolute extrema on a polygonal boundary watch

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    So I want to find the absolute extrema of the function:

     f(x,y) = 2x^2 - 3y^2 on  [-1,1] \times [-1,1] \subset \mathbb{R}^2

    So I did this:

     f_x = 4x, f_y = -6y \Rightarrow (0,0) is critical.

     f_{xx} = 4, f_{yy} = -6, f_{xy} = 0

     f_{xx}f_{yy} - f{xy}^2 = -24 < 0 so (0,0) is a saddle.

    Then I restricted to each boundary:

     x = 1, y = t, t \epsilon [-1,1]

     \Rightarrow f(x,y) = 2 - 3t^2

     f'(t) = -6t = 0 \Rightarrow t = 0

     \Rightarrow x = 1, y = 0 \Rightarrow f(x,y) = 2

    and then again for restricting to x = -1 and y = t  t \epsilon [-1,1] which again gives f(x,y) = 2.

    Now again for x=t and y=1 for t  \epsilon [-1,1].
    \Rightarrow f(0,1) = -3

    and again for x=t and y=-1 for t  \epsilon [-1,1]. which also gives f(0,-1) = -3.

    However I do not know how to show whether these are maxima or minima. Also apparently I am supposed to check  f(\pm 1, \pm1)

    I don't really understand what I am doing with this question tbh.

    Any help appreciated. xxx
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    (Original post by adie_raz)
    So I want to find the absolute extrema of the function:

     f(x,y) = 2x^2 - 3y^2 on  [-1,1] \times [-1,1] \subset \mathbb{R}^2

    So I did this:

     f_x = 4x, f_y = -6y \Rightarrow (0,0) is critical.

     f_{xx} = 4, f_{yy} = -6, f_{xy} = 0

     f_{xx}f_{yy} - f{xy}^2 = -24 < 0 so (0,0) is a saddle.

    Then I restricted to each boundary:

     x = 1, y = t, t \epsilon [-1,1]

     \Rightarrow f(x,y) = 2 - 3t^2

     f'(t) = -6t = 0 \Rightarrow t = 0

     \Rightarrow x = 1, y = 0 \Rightarrow f(x,y) = 2

    and then again for restricting to x = -1 and y = t  t \epsilon [-1,1] which again gives f(x,y) = 2.

    Now again for x=t and y=1 for t  \epsilon [-1,1].
    \Rightarrow f(0,1) = -3

    and again for x=t and y=-1 for t  \epsilon [-1,1]. which also gives f(0,-1) = -3.

    However I do not know how to show whether these are maxima or minima. Also apparently I am supposed to check  f(\pm 1, \pm1)

    I don't really understand what I am doing with this question tbh.

    Any help appreciated. xxx
    When the domain is closed and have boundaries (is compact set) and f is
    continuous here then f has minima and maxima here (Weierstrass)
    Where can be these?
    1. Within the domain
    - Where the function is not differentiable
    - Where has local extrema (you find a saddle)
    (The surface resembles a saddle that curves up in one direction,
    and curves down in a different direction). But this neither minima nor maxima
    2. At the boundary (a case of function with one variable)
    You checked the boundaries (fixing the proper x and y values).
    You found the extrema and you should order them to find
    minima and maxima
    (AS you can see x=1 and x=-1 gives f(1,0)=f(0,-1)=2 so from y direction the saddle
    is minima and the suface curves up, similarly for y=1 and y=-1 f(0,1)=f(0,-1)=-3
    that is the saddle from x direction is maxima and the surface curves down)
    Because the surface has saddle at (0,0) the extrema on the boundaries are
    the minima and maxima.
    Minimun at: (0,1 and (0,-1)
    Maximum at: (1,0) and (-1,0)
    These are absolute on the closed set.
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    (Original post by ztibor)
    When the domain is closed and have boundaries (is compact set) and f is
    continuous here then f has minima and maxima here (Weierstrass)
    Where can be these?
    1. Within the domain
    - Where the function is not differentiable
    - Where has local extrema (you find a saddle)
    (The surface resembles a saddle that curves up in one direction,
    and curves down in a different direction). But this neither minima nor maxima
    2. At the boundary (a case of function with one variable)
    You checked the boundaries (fixing the proper x and y values).
    You found the extrema and you should order them to find
    minima and maxima
    (AS you can see x=1 and x=-1 gives f(1,0)=f(0,-1)=2 so from y direction the saddle
    is minima and the suface curves up, similarly for y=1 and y=-1 f(0,1)=f(0,-1)=-3
    that is the saddle from x direction is maxima and the surface curves down)
    Because the surface has saddle at (0,0) the extrema on the boundaries are
    the minima and maxima.
    Minimun at: (0,1 and (0,-1)
    Maximum at: (1,0) and (-1,0)
    These are absolute on the closed set.
    Well why then do my notes say to check  f(\pm 1, \pm 1) which is the local minimum when  x = \pm 1 and the local maximum when  y = \pm 1 I do not understand why I must check these values and I am not sure I understand your explanation of "from the x direction", "from the y direction" is there not a way to check if something you have found is a maxima or minima such as second derivative when working in two dimensions...

    Edit: Looking at it  f''(t) = 4, -6 so I can see they are minimum and maximum respectively.

    Now I just want to know why I would check  f(\pm1,\pm1) at all...
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    (Original post by adie_raz)
    Well why then do my notes say to check  f(\pm 1, \pm 1) which is the local minimum when  x = \pm 1 and the local maximum when  y = \pm 1 I do not understand why I must check these values and I am not sure I understand your explanation of "from the x direction", "from the y direction" is there not a way to check if something you have found is a maxima or minima such as second derivative when working in two dimensions...
    That was only the geometric interpretation of the saddle and consequences of it. For example the cross section of the surface (palallel with (xz) plane passing through (0,0) I denoted it 'from x direction') will give a level curve (parabole) with minimum at x=0 (that is the saddle on surface), and with maximum of 2 when x=1 or x=-1 (those are on the boundaries). So f.e f(1,0)=2 will be maximum at the boundary. This maybe only a device to rank the local extrema of surface found on the boudaries.
    Edit: Looking at it  f''(t) = 4, -6 so I can see they are minimum and maximum respectively.

    Now I just want to know why I would check  f(\pm1,\pm1) at all...
    WHen you find the stationary points of a surface at above a compact set
    then according to criteria of Weierstrass it is needed to have the function maxima and minima at this set.
    1. These can be within the boundaries
    You can find them with second derivative test.
    2. And can be extrema at the boundaries

    Finding minima and maxima at boundaries is similar to above
    just for a function with one variable (fixing one of the two variable)
    So your surface boundaries are funtions with one variable over closed real intervals .
    Acccording to the Weierstrass the function has minima and maxima above this
    interval.
    These can be
    1. within the interval
    1. at the boundary of the interval that is at the starting and ending point
    YOu can find them with first derivative test. Maybe you find only one or none
    (maxima or minima) within the interval so you should to find the other or them
    at the strating or ending point of interval, because maxima and minima must exist somewhere at the closed interval.

    THese points are the vertice of the polygonal boundary in your question.
 
 
 
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