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Math Help - Maximum/Minimum

  1. #1
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    Maximum/Minimum

    It's all in the pdf.

    10x in advance.
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  2. #2
    MHF Contributor kalagota's Avatar
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    do you know the second derivative test? if so, you can do it..
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  3. #3
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    Quote Originally Posted by kalagota View Post
    do you know the second derivative test? if so, you can do it..
    What's that?
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by asi123 View Post
    What's that?
    If the second derivative is positive at a critical value then the critical value is a relative minimum. If the second derivative is negative then the critical value is a relative maximum.

    -Dan

    EDIT: Sorry, I hadn't looked at the actual problem. The test I'm referring to is for one dimensional Calculus. I'll let someone who knows the subject better take this one.
    Last edited by topsquark; July 12th 2008 at 07:45 AM. Reason: Addendum
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  5. #5
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    Quote Originally Posted by topsquark View Post
    If the second derivative is positive at a critical value then the critical value is a relative minimum. If the second derivative is negative then the critical value is a relative maximum.

    -Dan

    Ok, here is the second derivative, how can I know if it's positive or negative for a given point?

    What do I do with dx and dy?

    10x.
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Ok, I couldn't open your PDF and your image won't appear either.

    But here is what I think you are asking

    Let (a,b) denote a critical point of f(x,y)

    In other words f_x(a,b)=f_y(a,b)=0

    or f_x(a,b)\quad\text{or}\quad{f_y(a,b)} is undefined

    Now define

    D(a,b)=f_{xx}(a,b)f_{yy}(a,b)-\left\{f_{xy}(a,b)\right\}^2


    Now if D(a,b)>0\quad\text{and}\quad{f_{xx}(a,b)>0} then the point (a,b) is a relative min.

    If D(a,b)>0\quad\text{and}\quad{f_{xx}(a,b)<0} then (a,b) is a relative max.

    If D(a,b)<0 then (a,b) is a Saddle Point of f(x,y)

    And if D(a,b)=0 no conclusion can be drawn.


    Is this what you are asking about? If not just respond saying so.

    Mathstud.
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  7. #7
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    Quote Originally Posted by Mathstud28 View Post
    Ok, I couldn't open your PDF and your image won't appear either.

    But here is what I think you are asking

    Let (a,b) denote a critical point of f(x,y)

    In other words f_x(a,b)=f_y(a,b)=0

    or f_x(a,b)\quad\text{or}\quad{f_y(a,b)} is undefined

    Now define

    D(a,b)=f_{xx}(a,b)f_{yy}(a,b)-\left\{f_{xy}(a,b)\right\}^2


    Now if D(a,b)>0\quad\text{and}\quad{f_{xx}(a,b)>0} then the point (a,b) is a relative min.

    If D(a,b)>0\quad\text{and}\quad{f_{xx}(a,b)<0} then (a,b) is a relative max.

    If D(a,b)<0 then (a,b) is a Saddle Point of f(x,y)

    And if D(a,b)=0 no conclusion can be drawn.


    Is this what you are asking about? If not just respond saying so.

    Mathstud.

    I rescan everything in BMP so I hope you can see it now.

    What I'm talking about is a function with a condition, after I find all the points, how do I determinate which one is maximum and which one is minimum, I know I need to use d^2f somehow but I dont know how.

    10x in advance.
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by asi123 View Post
    I rescan everything in BMP so I hope you can see it now.

    What I'm talking about is a function with a condition, after I find all the points, how do I determinate which one is maximum and which one is minimum, I know I need to use d^2f somehow but I dont know how.

    10x in advance.
    Im really sorry, but once again I cannot see it. If you are not talking about finding relative extrema then I assume this is a question of either Absolute Extrema in a bounded region or a question of Lagrange Multipliers.

    Now this may sound novel, but have you tried just evaluating all the points, and from there since you know that they are all relative extrema you can seee which are the absolute max and mins?
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  9. #9
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    Quote Originally Posted by Mathstud28 View Post
    Im really sorry, but once again I cannot see it. If you are not talking about finding relative extrema then I assume this is a question of either Absolute Extrema in a bounded region or a question of Lagrange Multipliers.

    Now this may sound novel, but have you tried just evaluating all the points, and from there since you know that they are all relative extrema you can seee which are the absolute max and mins?

    You right, I'm talking about Lagrange Multipliers, lets say I have a function f(x,y) and also a condition g(x,y), and I want to build a function l(x,y) = f(x,y) + gama * g(x,y). From here I find all the points, my question is how can I tell which one is minimum and which one is maximum? can I just put them in f(x,y) and evaluate it from there? shouldn't I use d^2l or something like that?

    10x again.
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  10. #10
    MHF Contributor Mathstud28's Avatar
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    Ok, so I finally see your page it says

    Maximize f(x,y)=x^2+12xy+2y^2

    Under the condition that

    4x^2+y^2=25

    So defining

    g(x,y)=4x^2+y^2-25

    We need to solve the following system of equations

    \left\{\begin{array}{cc}f(x,y)_x=\lambda{g(x,y)_x}  \\f(x,y)_y=\lambda{g(x,y)_y}\\g(x,y)=0\end{array}\  right\}

    So computing

    f_x=2x+12y

    f_y=12x+4y

    g_x=8x

    g_y=2y

    We see that we must solve

    \left\{\begin{array}{cc}2x+12y=8\lambda{x}\\12x+4y  =2\lambda{y}\\4x^2+y^2-25=0\end{array}\right\}

    So we see that the four solutions are

    \left\{\begin{array}{c}x=-2\\y=3\\z=-2\end{array}\right\}

    and

    \left\{\begin{array}{c}x=\frac{-3}{2}\\y=-4\\z=\frac{17}{4}\end{array}\right\}

    and

    \left\{\begin{array}{c}x=\frac{3}{2}\\y=4\\z=\frac  {17}{4}\end{array}\right\}

    and

    \left\{\begin{array}{c}x=2\\y=-3\\z=-2\end{array}\right\}


    So testing we get

    f(-2,3)=-50

    f\left(\frac{-3}{2},-4\right)=\frac{425}{4}

    f\left(\frac{3}{2},4\right)=\frac{-151}{4}

    f(2,-3)=-50

    You can draw your conclusions from there.
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