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Thread: Absolute Extrema (two variables)

  1. #1
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    Absolute Extrema (two variables)

    I need a little help with finding absolute extrema bounded by a region.

    Find the absolute extrema of $\displaystyle G(x,y) = xy$ over the region $\displaystyle 4x^2+y^2\leq2$.

    $\displaystyle G_{x}= y$
    $\displaystyle G_{y}= x$

    Therefore:

    $\displaystyle y=0$ and $\displaystyle x=0$

    $\displaystyle G(0,0) = 0$

    Where do I go from here and how?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by wilcofan3 View Post
    I need a little help with finding absolute extrema bounded by a region.

    Find the absolute extrema of $\displaystyle G(x,y) = xy$ over the region $\displaystyle 4x^2+y^2\leq2$.

    $\displaystyle G_{x}= y$
    $\displaystyle G_{y}= x$

    Therefore:

    $\displaystyle y=0$ and $\displaystyle x=0$

    $\displaystyle G(0,0) = 0$

    Where do I go from here and how?
    now find the extrema on the boundary $\displaystyle 4x^2 + y^2 = 2$

    solve for $\displaystyle y$, say, and plug that expression in $\displaystyle G(x,y)$, to get a new single variable function $\displaystyle G(x,x)$. then find its (absolute) extreme points

    you can then find the absolute extrema after finding all extreme points
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  3. #3
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    By the Extreme Value Theorem, the extremes of $\displaystyle G(x, y) = xy$ given the constraint $\displaystyle 4x^2 + y^2 \leq 2$ are either at points where $\displaystyle G_x = G_y = 0$ or on the border of the region (i.e. where $\displaystyle 4x^2 + y^2 = 2$, since this is an ellipse). Let $\displaystyle H(x, y) = 4x^2 + y^2$. Using the method of Lagrange Multipliers,

    $\displaystyle G_x = \lambda H_x$
    $\displaystyle G_y = \lambda H_y$

    $\displaystyle y = \lambda \cdot 8x$
    $\displaystyle x = \lambda \cdot 2y$

    $\displaystyle \lambda = \frac{y}{8x} = \frac{x}{2y}$

    $\displaystyle 2y^2 = 8x^2$

    $\displaystyle y = \pm 2x$

    Substituting back into $\displaystyle H(x, y) = 2$

    we have
    $\displaystyle 4x^2 + (2x)^2 = 2$
    $\displaystyle 4x^2 + 4x^2 = 2$
    $\displaystyle 8x^2 = 2$
    $\displaystyle 4x^2 = 1$
    $\displaystyle x^2 = \frac{1}{4}$

    $\displaystyle x = \pm \frac{1}{2}$

    $\displaystyle y = \pm 1$

    Hence the minimum value of $\displaystyle G(x, y) = xy$ is $\displaystyle -\frac{1}{2}$
    and the maximum value of $\displaystyle G(x, y) = xy$ is $\displaystyle \frac{1}{2}$.
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