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=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
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
Let's assume that the length of the property is 'L' yards and the width is 'W' yards.
We are given that 200 yards of fencing will enclose three sides of the property, which means the perimeter of the rectangle (excluding the side adjacent to the river) is 200 yards.
Perimeter of a rectangle = 2(length + width)
Given that the perimeter is 200 yards:
2(L + W) = 200
Divide both sides of the equation by 2:
L + W = 100
Also, we are given that the area of the property is 3200 square yards:
Area of a rectangle = length x width
Given that the area is 3200 square yards:
L x W = 3200
Now, we have a system of equations:
Equation 1: L + W = 100
Equation 2: L x W = 3200
From Equation 1, we can solve for one variable in terms of the other:
L = 100 - W
Substitute this value of L in Equation 2:
(100 - W) x W = 3200
Expand:
100W - W^2 = 3200
Rearrange the equation in standard quadratic form:
W^2 - 100W + 3200 = 0
Now, we can solve this quadratic equation using factoring, completing the square, or the quadratic formula. To simplify the process, let's use factoring:
(W - 80)(W - 40) = 0
This gives us two possible values for W:
W = 80 or W = 40
If W = 80, then from Equation 1:
L + 80 = 100
L = 20
If W = 40, then from Equation 1:
L + 40 = 100
L = 60
Therefore, the possible dimensions of the property are:
1) Length = 20 yards and Width = 80 yards
2) Length = 60 yards and Width = 40 yards
To find the dimensions of the rectangular property, we can start by considering the information given.
We know that the farmer has 200 yards of fencing, which will be used to enclose three sides of the rectangle. Since the fourth side will be the river, we don't need to account for fencing there.
Let's denote the width of the rectangular property as "w" and the length as "l", both measured in yards.
The perimeter of the rectangle will be the sum of the lengths of the three sides that require fencing. Therefore, we have:
Perimeter = 2w + l = 200 yards
We're also given that the area of the property is 3200 square yards. The area of a rectangle is calculated by multiplying its length by its width. Thus, we have:
Area = length * width = l * w = 3200 square yards
We now have a system of two equations with two variables that we can solve.
Starting with the perimeter equation, we can isolate "l":
2w + l = 200
l = 200 - 2w
Substituting this expression for "l" into the area equation:
3200 = (200 - 2w) * w
Expanding and rearranging the equation:
3200 = 200w - 2w^2
Rearranging the equation to a quadratic form:
2w^2 - 200w + 3200 = 0
To solve this quadratic equation, we can use factoring, completing the square, or the quadratic formula. In this case, factoring is the most straightforward method.
Factoring the quadratic equation, we get:
2w^2 - 200w + 3200 = 0
2(w^2 - 100w + 1600) = 0
2(w - 40)(w - 40) = 0
This gives us two possibilities: w - 40 = 0, which leads to a negative value for width, and w - 40 = 0, which implies w = 40 yards.
Substituting this value back into the perimeter equation:
2w + l = 200
2(40) + l = 200
80 + l = 200
l = 200 - 80
l = 120 yards
Therefore, the width of the property is 40 yards and the length is 120 yards.