# 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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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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