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Thread: algebra probs

  1. #1
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    algebra probs

    Problem:
    $\displaystyle f(x,) = x^2+x(y^2-1)$

    a) find the extremes and classify them.
    b) now, $\displaystyle x^2+x^2 \leq 17$ Calculate the highest and lowest f(x,y)

    Skärmavbild 2012-09-11 kl. 19.42.48.png


    Attempt:
    a)
    2012-09-11 19.39.20.jpg
    b)
    2012-09-11 19.39.50.jpg


    The thing is this: the anser is that x = -√17 is also an alternative. But I got x = 8/3 och x = -2.
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  2. #2
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    Re: algebra probs

    Quote Originally Posted by jacob93 View Post
    Problem:
    $\displaystyle f(x,) = x^2+x(y^2-1)$

    a) find the extremes and classify them.
    b) now, $\displaystyle x^2+x^2 \leq 17$ Calculate the highest and lowest f(x,y)

    Skärmavbild 2012-09-11 kl. 19.42.48.png


    Attempt:
    a)
    2012-09-11 19.39.20.jpg
    b)
    2012-09-11 19.39.50.jpg


    The thing is this: the anser is that x = -√17 is also an alternative. But I got x = 8/3 och x = -2.
    We can use the second derivative test to find the extreem values.

    $\displaystyle \frac{\partial f}{\partial x}=2x+(y^2-1)=0$

    and

    $\displaystyle \frac{\partial f}{\partial y}=2xy=0$

    The 2nd equation gives rise to two different cases x=0 and y=0.

    Case I: if x=0 then

    $\displaystyle y^2-1=0 \implies y=\pm1$

    Case II: If y=0 then,

    $\displaystyle 2x-1=0 \implies x=\frac{1}{2}$

    Now we can make the Hessian matrix

    $\displaystyle H(x,y)=\begin{vmatrix}f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{vmatrix}=\begin{vmatrix}2 & 2y \\ 2y & 2x \end{vmatrix}=4x-4y^2$

    Now we can check the critical points found above

    Case I: $\displaystyle (0,\pm 1)$

    $\displaystyle H(0, \pm 1)=4(0)-4(\pm 1)^2=-4$

    So both of these are Saddle points.

    Case I: $\displaystyle \left(\frac{1}{2}, 0 \right)$

    $\displaystyle H\left( \frac{1}{2},0 \right)=4\cdot \frac{1}{2}-4(0)^2=2$

    Since both eigen values of the Hessian are positive the point in a local minimum.

    For Part b) all we need to do is calculate the maximum and minimum on the boundary.

    Since the boundary is a circle with radius $\displaystyle \sqrt{17}$, it can be parameterized by

    $\displaystyle x=\sqrt{17}\cos(t) \quad y=\sqrt{17}\sin(t)$

    If we put these into the original function we get

    $\displaystyle f(t) = 17 \cos^2(t)-\sqrt{17}\cos(t)(17\sin^2(t)-1)$

    This is a function of one variable and its extreem values can be found using methods from first semester calculus.

    Work not shown, but you will find that $\displaystyle t=\pi n, \quad n \in \mathbb{Z}$ is a critical point. If you put this back into the parameteric equations you get

    $\displaystyle x=\sqrt{17}\cos(\pi n)=\pm \sqrt{17} \quad y=\sqrt{17}\sin(\pi n)=0$.
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  3. #3
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    Re: algebra probs

    Forgot the boundary, thanks.
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