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a wire is 36 m long is cut into two pieces , each piece is bent to form a rectangle which is 1 centimeter longer than its width. how long should each piece be to minimize the sum of the areas of the two rectangles?

  • math - ,

    seems like I saw this one go by the other day, but here it is again:

    If the two pieces are x and y in length, then the rectangles have dimensions

    (x-2)/4 by (x-2)/4 + 1 or
    (x-2)/4 by (x+2)/4
    (y-2)/4 by (y+2)/4

    But, y = 36-x, so the rectangles are

    (x-2)/4 by (x+2)/4
    (34-x)/4 by (38-x)/4

    and the sum of their areas is

    (x-2)(x+2)/16 + (34-x)(38-x)/16
    = (x^2-36x+644)/8
    = 1/8 [(x-18)^2 + 320]

    This is a parabola with vertex at (18,40), where the area is a minimum.

  • or - math - ,

    avoiding some of those fractions

    let the sides of one rectangle be x and x+1
    and the other, y and y+1
    4x+2 + 4y+2 = 36
    y = 8-x

    area = A = x(x+1) + y(y+1)
    = x^2 + x + y^2 + y
    = x^2 + x + 64-16x+x^2 + 8-x
    = 2x^2 - 16x + 72
    Quadratic opening up, so a minimum
    dA/dx = 4x - 16
    = 0 for a min
    x = 4
    then y = 4

    The two rectangles are the same size, both 4 by 5
    so we need 2(4+5) or 18 m for each
    for a min area of 4(5) + 4(5) or 40 m^2
    for a min

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