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Calc-optimization

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Given y=(x)^1/2, find the closest point to (3/2,0)

  • Calc-optimization -

    let the closest point be P(a,b)
    label the given point A(3/2 , 0)

    AP will have a minimum length when AP is perpendicular to the tangent at P

    slope of tangent = y' = (1/2)x^(-1/2) or 1/(2√x)
    so at P the slope of the tangent is 1/2√a

    slope of AP = b/(a-3/2)

    b/(a-3/2) = -2√a , (the negative reciprocal of 1/2√a
    but also b = √a

    √a/(a-3/2) = -2√a
    1/(a-3/2) = -2
    1 = -2a + 3
    2a = 2
    a = 1
    then b=√1 = 1

    the closest point is (1,1)

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