math
posted by Amber .
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

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=2002y Ya
for (i) (2002y)*y=3200 Ya^2
200y2y^2=3200 seconddegree 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
first restriction:
2x + y = 200 > y = 2002x
Second piece of data:
xy = 3200
x(2002x) = 3200
200x  2x^2  3200 = 0
2x^2  200x + 10000 = 100003200
x^2  100x = 1600
x^2  100x + 2500 = 25001600
(x50)^2 = 900
x50 = ± 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