Posted by Amber on Wednesday, September 5, 2012 at 2:26am.
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
let the length of the field by y yds, (parallel to river)
leth the width of the river be x yds
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
40 + 2(80) = 200
80(40) = 3200
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