math

a farmer has 200 yards of fencing to enclose three sides of a rectangular piece of property that lies next to a river. the river will serve as the fourth side. find the dimensions of property if the area is 3200 yards^2

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asked by Amber
  1. i)the area of rectangle is x*y (whit x,y side)=>x*y=3200 Ya^2

    ii)the condition imposed from the questions is: x+y+y=200 Ya (i have choise two y because the text not exsplicit the misure of one the side):
    x+2y=200 => x=200-2y Ya

    for (i) (200-2y)*y=3200 Ya^2
    200y-2y^2=3200 second-degree equation
    you know resolve

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  2. let the length of the field by y yds, (parallel to river)
    leth the width of the river be x yds

    first restriction:
    2x + y = 200 ----> y = 200-2x

    Second piece of data:
    xy = 3200
    x(200-2x) = 3200
    200x - 2x^2 - 3200 = 0
    2x^2 - 200x + 10000 = 10000-3200
    x^2 - 100x = 1600
    x^2 - 100x + 2500 = 2500-1600
    (x-50)^2 = 900
    x-50 = ± 30
    x = 80 or a negative, which will be rejected

    the width is 80 yds and the length is 40

    check:
    40 + 2(80) = 200
    80(40) = 3200

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    posted by Reiny

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