calculus

A long rectangular sheet of metal, 12 inches wide, is to be made into a rain gutter by turning up two sides so that they are perpendicular to the sheet. How many inches should be turned up to give the gutter its greatest capacity?

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  1. All we have to consider is the area of the cross-section, since the length is a constant.

    let the sides to be turned up at both ends be x inches long
    let the remaining base be y inches
    2x + y = 12
    y = 12-2x

    area = xy = x(12-2x) = 12x - 2x^2
    d(area)/dx = 12 - 4x = 0 for a max of area
    4x = 12
    x = 3

    So the turn-ups should be 3 inches

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  2. Greatest capacity will be achieved when the area of cross section will be maximum.
    Let X be the base ,so (12-X)/2 will be the height.
    Area= Base x height= X(12-X)/2
    =(12X-X^2)/2
    Following the maximum theory, we differentiate the area w.r.t X and equate it to zero.
    d/dX of(Area)=(12-2X)/2 =0
    or 6-X=0
    or X=6 inches(base)
    so, height= (12-6)/2 = 3 inches

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  3. x (2x-2)
    d/dX of(Area)=(12-2X)/2 =0
    or 6-X=0
    or X=6 inches(base)
    so, height= (12-6)/2 = 3 inches

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  4. 7inches long 5 inches wide

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