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Thread: need help

  1. #1
    Senior Member
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    need help

    Determine the maximum and minimum of the function

    in the square having corners at points and .
    my progress:

    derivate on x i get $\displaystyle 12x^2-6y$
    derivate on y i get $\displaystyle 12y-6x$
    idk if this is right thinking but i wanna solve x and i get
    $\displaystyle y=6x/12$
    so i got $\displaystyle 12x^2-6x/12=0$ and we get that x_1=0 and x_2=1/24 i am right or wrong and what shall i do next?

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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: need help

    You have the right idea. You want to equate the partials to zero and solve the resulting system to find the critical points:

    $\displaystyle f_x(x,y)=12x^2-6y=0$

    $\displaystyle f_y(x,y)=12y-6x=0$

    Solve this system, we find:

    $\displaystyle (x,y)=(0,0),\,\left(\frac{1}{4},\frac{1}{8} \right)$

    Now you want to use the second partials test for relative extrema:

    Let $\displaystyle (a,b)$ be a critical point of $\displaystyle z=f(x,y)$ and suppose $\displaystyle f_{xx},\,f_{yy}$ and $\displaystyle f_{xy}$ are continuous in a rectangular region containing $\displaystyle (a,b)$.

    Let $\displaystyle D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-(f_{xy}(x,y))^2$.

    i) If $\displaystyle D(a,b)>0$ and $\displaystyle f_{xx}(a,b)>0$, then $\displaystyle f(a,b)$ is a relative minimum.

    ii) If $\displaystyle D(a,b)>0$ and $\displaystyle f_{xx}(a,b)<0$, then $\displaystyle f(a,b)$ is a relative maximum.

    iii) If $\displaystyle D(a,b)<0$, then $\displaystyle f(a,b)$ is not an extremum.

    iv) If $\displaystyle D(a,b)=0$, then no conclusion can be drawn concerning a relative extremum.

    Using this test, what conclusions can be drawn about the critical points?
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