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28 January, 02:09

The length of a rectangle is 2 cm less than 6 times the width. find the possibilities for the width of the rectangle such that the area is not more than 840 cm2

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  1. 28 January, 02:17
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    To solve this problem, let us first assign the variables. Let us say that:

    l = length of rectangle

    w = width of rectangle

    The given problem gives us the relation that:

    l = 6 w - 2 - - - > 1

    We know that the formula for area of rectangle is given as:

    A = l w - - - > 2

    Substituting equation 1 into 2:

    A = (6 w - 2) w

    A = 6 w^2 - 2 w

    The area must not be greater 840 cm^2, therefore:

    840 cm^2 = 6 w^2 - 2 w

    w^2 - (1/3) w = 140

    By completing the square:

    w^2 - (1/3) w + (1/36) = 140 + (1/36)

    (w - 1/6) ^2 = 5041/36

    w - 1/6 = ± 11.83

    w = - 11.66, 12

    Therefore the maximum value that the width can take is 12 cm. Therefore the possibilities of width such that the area does not exceed 840 cm^2 is from 1 to 12 cm.

    Answer:

    1 cm to 12 cm
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