A rancher wants to fence in an area of 2000000 square feet in a rectangular field and then divide it in half with a fence down the middle parallel to one side. What is the shortest length of fence that the rancher can use?
To find the shortest length of fence that the rancher can use, we need to determine the dimensions of the rectangular field.
Let's assume the dimensions of the rectangular field are as follows:
Length = L
Width = W
According to the problem, the rectangular field has an area of 2000000 square feet. Therefore, we can set up the equation:
L * W = 2000000
Now, since the rancher wants to divide the field in half with a fence down the middle parallel to one side, it means that the length of the field will be divided into two equal parts. Let's assume the width of the field (W) remains the same and the length (L) is divided into two equal parts, each of length L/2.
Now, we can substitute L/2 for L in the equation:
(W) * (L/2) = 2000000
Simplifying the equation:
WL/2 = 2000000
WL = 2 * 2000000
WL = 4000000
Now, we know that the total length of fence needed is the sum of all four sides of the rectangular field plus the fence down the middle. The formula for the perimeter of a rectangle is:
Perimeter = 2 * (Length + Width)
Since we divided the length (L) into two equal parts (L/2), we can calculate the perimeter as follows:
Perimeter = 2 * [(L/2) + W + (L/2)]
Simplifying the equation:
Perimeter = L + W + L
Perimeter = 2L + W
We can substitute WL = 4000000 into the equation:
Perimeter = 2(4000000/W) + W
To find the shortest length of fence, we need to minimize the value of the perimeter by taking its derivative with respect to W and set it equal to zero. Solving this equation will help us find the optimal width that minimizes the perimeter, thus giving us the shortest length of fence.
Differentiating the equation with respect to W gives:
d(Perimeter)/dW = -8000000/W^2 + 1
Setting d(Perimeter)/dW equal to zero and solving for W:
-8000000/W^2 + 1 = 0
-8000000 = W^2
W^2 = 8000000
W ≈ 2828.43 (rounded to two decimal places)
Now that we have the optimal width (W ≈ 2828.43), we can substitute it back into the perimeter equation to find the shortest length of fence:
Perimeter = 2(4000000/W) + W
Perimeter ≈ 2828.43 + 2(4000000/2828.43)
Perimeter ≈ 2828.43 + 2828.43 ≈ 5656.86
Therefore, the shortest length of fence that the rancher can use is approximately 5656.86 feet.
To find the shortest length of fence, we need to determine the dimensions of the rectangular field first.
Let's assume the width of the rectangular field is x.
Since the area of a rectangle is given by length × width, we can set up the equation:
length × width = area
x × width = 2000000
Solving for width, we can rearrange the equation as:
width = area / length
width = 2000000 / x
To find the minimum amount of fence required, we want to minimize the length of fence used. This occurs when the length and width of the rectangular field are equal, which also means the field is a square.
So, we can set x = width. Substituting this value in the earlier equation:
x × x = 2000000
x^2 = 2000000
Taking the square root of both sides:
√(x^2) = √2000000
x = √2000000
Hence, the width and length of the rectangular field are both √2000000.
Now, to calculate the total length of fence required, we need to account for the fence down the middle that divides the field into two equal halves. We can assume this fence is of length x, given that the width and length of the field are equal.
Therefore, the shortest length of fence that the rancher can use is:
2 × (width + length + fence down the middle)
= 2 × (√2000000 + √2000000 + √2000000)
= 2 × 3 × √2000000
≈ 6 × √2000000
To simplify the square root, we can express it as the product of a perfect square and the remaining square root part:
6 × √2000000 = 6 × √(100 × 20000) = 6 × √(10^2 × 20000)
Using the property √(a × b) = √a × √b, we can simplify further:
6 × √(10^2 × 20000) = 6 × 10 × √20000 = 60 × √20000
Hence, the shortest length of fence that the rancher can use is approximately 60 times the square root of 20000.
fence length = z = 3 x + 2 y
x y =2*10^6 ft^2
so
z = 3(2*10^6/y) + 2 y
z = 6*10^6/y + 2y
dz/dy = -6*10^6/y^2 + 2 y
= 0 for max or min
y^3 = 3*10^6
y = 3^(1/3) *100
y = 144 ft
x = 13889 ft
need length = 3x+2y