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22 December, 03:39

Find the dimensions of the open rectangular box of maximum volume that can be made from a sheet of cardboard 41 in. by 22 in. by cutting congruent squares from the corners and folding up the sides. Then find the volume.

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  1. 22 December, 05:09
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    V (max) = 1872.42 in³

    Step-by-step explanation:

    Let call " x " the side of the cut square in each corner, then the sides of rectangular base (L and W) will be

    L = 41 - 2*x and

    W = 22 - 2*x x = the height of the open box

    Volume of the box is

    V (b) = L*W*x

    And volume of the box as a function of x is:

    V (x) = (41 - 2*x) * (22 - 2*x) * x

    V (x) = (902 - 82*x - 44*x + 4*x²) * x ⇒ V (x) = (902 - 126*x + 4x²) * x

    V (x) = 902*x - 126*x² + 4x³

    Taking derivatives on both sides of the equation we get:

    V' (x) = 902 - 252*x + 12 * x²

    V' (x) = 12*x² - 252*x + 902

    V' (x) = 0 ⇒ 12*x² - 252*x + 902 = 0

    A second degree equation solving for x (dividing by 2)

    6*x² - 126 * x + 451 = 0

    x₁,₂ = (126 ± √ 15876 - 10824) / 12

    x₁,₂ = (126 ± 71,08) / 12

    x₁ = 16 in (we dismiss this value since 2x > than 22 there is not feasible solution for the problem)

    x₂ = 54.92/12 ⇒ x₂ = 4,58 in

    Then maximum volume is:

    V (max) = 31.84 * 12.84 * 4,58

    V (max) = 1872.42 in³
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