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4 January, 11:23

An open rectangular box with square base is to be made from 48 ft^2 of material. What dimensions will result in a box with the largest possible volume?

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  1. 4 January, 11:33
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    Are of bottom is : x^2+4

    sides 4*xy

    Surface area = 48ft^2 = x^2 + 4xy

    volume is x^2*y

    V = x^2*y
  2. 4 January, 11:43
    0
    If we have a square base; then:

    long=width=x

    height = h

    Area of this open rectangular box=x²+4 (xh)

    then:

    x²+4xh=48

    4xh=48-x²

    h = (48-x²) / 4x

    let:

    V (x, h) = volume of this open rectangular box

    V (x, h) = x²h

    V (x) = x² (48-x²) / 4x

    V (x) = x (48-x²) / 4

    V (x) = (48x-x³) / 4

    1) we have to find the first derivative of V (x):

    V' (x) = (48-3x²) / 4

    2) We find out the values of "x", When V' (x) = 0

    (48-3x²) / 4=0

    48-3x²=0*4

    48-3x²=0

    -3x²=-48

    x²=-48/-3

    x²=16

    x=⁺₊√16

    We have two possible solutions:

    x=-4; this is not solution, because the long or width can't be negative.

    x=4; it is possible solutions

    3) we have to find the second derivative

    V'' (x) = - 6x/4=-3x/2

    V'' (4) = - 12/2=-6<0; then we have a maximum at x=4;

    4) we find the value of "h", when x=4

    h = (48-x²) / 4x

    h = (48-4²) / 4 (4)

    h = (48-16) / 16

    h=32/16

    h=2

    Therefore:

    long=width=4 ft

    height=2 ft

    V (x, y) = x²h = (4 ft) ² (2ft) = 32 ft³

    Answer: the largest possible volume would be 32 ft³
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