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Math Help - More Optimization Problems . . .

  1. #1
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    More Optimization Problems . . .

    I need some help with this problem.

    Find the points on the ellipse 4x^2 + y^2 = 4 that are farthest away from the point (1, 0).

    d = sqrt[(x -1)^2 + (y - 0)^2]

    4x^2 + y^2 = 1
    => x^2 + y^2/4 = 1
    => x^2 = 1 - y^2/4
    => x = sqrt(1 - y^2/4)
    => x = 1 - y/2 = -y

    d = sqrt[(-y -1)^2 + (y - 0)^2]
    d^2 = f(y) = (-y -1)^2 + (y -0)^2
    f'(y) = 2(-y -1) + 2(y - 0) = -2y -2 + 2y - 0 = -2

    I don't know what to do next, and I'm sure I've made a couple of mistakes anyway.

    Thanks.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by zachb View Post
    I need some help with this problem.

    Find the points on the ellipse 4x^2 + y^2 = 4 that are farthest away from the point (1, 0).

    d = sqrt[(x -1)^2 + (y - 0)^2]
    First, if you maximise d you also maximise d^2, using d^2 avoids that
    sqrt so makes everything easier, so we want to maximise:

    D2 = d^2 = (x-1)^2 + (y-0)^2 = (x-1)^2 + y^2

    but from the equation for the ellipse:

    y^2 = 4 - 4x^2

    So:

    D2 = (x-1)^2 - 4(x^2 -1) = - 3 x^2 - 2 x + 5

    Now:

    d(D2)/dx = -6 x -2

    for a maximum of D2 d(D2)/dx=0, so x=-1/3, and so D2 = 16/3, and y=4sqrt(2)/3
    (assuming I have done the algebra correctly).

    RonL
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  3. #3
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    You did the algebra correctly. Thanks. Solving for y^2 and them plugging it in actually made the problem a lot easier.
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