calculus

Find the point P in the first quadrant on the curve y=x^-2 such that a rectangle with sides on the coordinate axes and a vertex at P has the smallest possible perimeter.

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  1. let the point be P(x,y)

    Perimeter = 2x + 2y
    = 2x + 2(x^-2)
    = 2x + 2/x^2

    d(perimeter)/dx = 2 -4/x^3
    = 0 for min of perimeter

    2 - 4/x^3 = 0
    2 = 4/x^3
    x^3 = 2
    x = 2^(1/3) , (the cube root of 2)
    y = 2^(-2/3)

    (I really expected the point to be (1,1) but the Calculus shows my intuition was wrong
    the perimeter with the above answers is 3.78
    had it been (1,1) the perimeter would have been 4)

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    posted by Reiny
  2. Try this way:
    Perimeter = 2x+2

    therefore P = 2x + 2/x^2
    d(perimeter)/dx = (2x^3-4)/x^3
    solving for x = 2^(1/3)
    therefor y was given as y = x^-2
    plug x into y to solve for y.
    y = 2^(-2/3) or y = (1/4)^(1/3)
    check the answer it should be 3.78

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    posted by Rob

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