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Math Help - 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 G(x,y) = xy over the region 4x^2+y^2\leq2.

    G_{x}= y
    G_{y}= x

    Therefore:

    y=0 and x=0

    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 G(x,y) = xy over the region 4x^2+y^2\leq2.

    G_{x}= y
    G_{y}= x

    Therefore:

    y=0 and x=0

    G(0,0) = 0

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

    solve for y, say, and plug that expression in G(x,y), to get a new single variable function 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 G(x, y) = xy given the constraint 4x^2 + y^2 \leq 2 are either at points where G_x = G_y = 0 or on the border of the region (i.e. where 4x^2 + y^2 = 2, since this is an ellipse). Let H(x, y) = 4x^2 + y^2. Using the method of Lagrange Multipliers,

    G_x = \lambda H_x
    G_y = \lambda H_y

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

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

    2y^2 = 8x^2

    y = \pm 2x

    Substituting back into H(x, y) = 2

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

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

    y = \pm 1

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