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1 February, 16:54

Find the dimensions of the open rectangular box of maximum volume that can be made from a sheet of cardboard 29 in. by 16 in. by cutting congruent squares from the corners and folding up the sides. Then find the volume.

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  1. 1 February, 17:19
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    height = 3.3 in

    V = 694.848 in^3

    Step-by-step explanation:

    length of cardboard, l = 29 in

    width of cardboard, w = 16 in

    let the side of square is d.

    so the length of box = 29 - 2y

    width of box = 16 - 2y

    height of box = y

    Volume of box = length x width x height

    V = (29 - 2y) x (16 - 2y) x y

    V = 464 y - 90 y^2 + 4 y^3

    dV / dy = 464 - 180 y + 12y^2

    For maxima and minima, dV/dy = 0

    12y^2 - 180 y + 464 = 0

    By solving

    y = 3.3 in, 11.7 in

    Now find double differentiation

    d²V/dy² = 24 y - 180

    Put, y = 3.3 in, it is - 100.8

    Put, y = 11.7 in, it is + 100.8

    So for maximum volume, y = 3.3 in

    And the value of maximum volume

    V = 464 (3.3) - 90 (3.3 x 3.3) + 4 (3.3 x 3.3 x 3.3) = 1531.2 - 980.1 + 143.748

    V = 694.848 in^3
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