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25 January, 18:08

A company plans to enclose three parallel rectangular areas for sorting returned goods. the three areas are within one large rectangular area and 10641064 yd of fencing is available. what is the largest total area that can be enclosed?

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  1. 25 January, 18:19
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    First get the perimeter

    Perimeter: 2l + 2w = 1064 yd

    The region inside the fence is the area

    Area: A = lw

    We need to solve the perimeter formula for either length or width.

    2l + 2w = 1064 yd

    2w = 1064 yd - 2l

    W = 1064 - 2l / 2

    W = 532 - l

    Now substitute than to the area formula

    A = lw

    A = l (532 - l)

    A = 532 - l^2

    Since the equation A represents a quadratic expression, rewritte the expression with the exponents in descending order

    A (l) = - l^2 + 532l

    Then look for the value of the x coordinate

    l = - b/2a

    l = - 532/2 (-1)

    l = - 532/-2

    l = 266 yards

    Plugging in the value into our calculation for area:

    A (l) = - l^2 + 532

    A (266) = - (266) ^2 + 532 (266)

    A (266) = 70756 + 141512

    = 70756 square yards.

    Thus the largest area that could encompass would be a square where each side has a length of 266 yards and a width of:

    W = 532 - l

    = 532 - 266

    = 266
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