A total of 800 m of fencing material is used to enclose a rectangular field and to divide the field into four portions by fences parallel to one of the sides of the field. Find the maximum area which can be enclosed in this rectangle and it’s dimensions
as always in problems like this, the maximum area is achieved when the fencing is divided equally among lengths and widths.
Since you have 5 lengths and 2 widths, the field is 200x80
To go through the math, you have
5x+2y = 800
a = xy = x(800-5x)/2 = 400x - 5/2 x^2
The vertex of this parabola is at x = -b/2a = 400/5 = 80
Well, this sounds like quite the fencing challenge! Let's put on our clown hats and figure it out.
To maximize the area, we need to find the dimensions of the rectangle that will give us the largest possible area. Let's call the length of the field x and the width y.
We know that the total perimeter of the rectangle is 800 m, which means that the sum of all sides is 800 m. So, we have:
2x + 3y = 800
Now, let's solve this equation for one variable. Let's solve for x:
2x = 800 - 3y
x = (800 - 3y) / 2
Now, we can substitute this value of x into the formula for area (A = x * y):
A = [(800 - 3y) / 2] * y
Simplifying further, we get:
A = (800y - 3y^2) / 2
To find the maximum area, we need to find the vertex of this quadratic equation. We can use calculus or complete the square, but since we're clowning around, let's use calculus!
Taking the derivative of the equation with respect to y, we get:
dA/dy = 800/2 - 6y/2
dA/dy = 400 - 3y
Setting this derivative equal to zero, we find:
400 - 3y = 0
3y = 400
y = 400/3
Substituting this value back into the equation for x, we get:
x = (800 - 3(400/3)) / 2
x = (800 - 400) / 2
x = 400 / 2
x = 200
So, the dimensions of the rectangle that will maximize the area are x = 200 m and y = 400/3 m. Now, let's calculate the maximum area:
A = (800(400/3) - 3(400/3)^2) / 2
A = (320000/3 - 160000/3) / 2
A = 160000/3 / 2
A ≈ 88888.89 m²
So, the maximum area that can be enclosed in this rectangle is approximately 88888.89 square meters, with dimensions of 200 m by 400/3 m.
And there you have it! A circus-worthy answer to your fencing puzzle.
Let's assume the length of the rectangle is L and the width is W.
To enclose the rectangular field with 800 m of fencing material, we need to consider:
2 sides of length L
1 side of length W
2 sides for dividing the field (parallel to W)
This gives us the equation: 2L + W + 2W = 800
Simplifying this equation, we get: 2L + 3W = 800
Now, let's solve for L in terms of W:
2L = 800 - 3W
L = (800 - 3W) / 2
To find the maximum area, we need to maximize the product of L and W. Therefore, we need to substitute the value of L in terms of W into the area formula (A = L * W).
A = ((800 - 3W) / 2) * W
Next, we can simplify the equation by multiplying both sides by 2:
2A = (800 - 3W) * W
Expanding the equation:
2A = 800W - 3W^2
To find the maximum area, we need to find the value of W that maximizes this quadratic equation.
To do this, we can set the derivative of 2A with respect to W equal to zero and solve for W:
d(2A)/dW = 800 - 6W = 0
Solving for W, we get:
6W = 800
W = 800 / 6
W = 133.33 m (approximately)
Now, substitute this value of W back into the equation for L:
L = (800 - 3(133.33)) / 2
L = (800 - 400) / 2
L = 200 m
Therefore, the maximum area that can be enclosed in this rectangle is:
A = L * W
A = 200m * 133.33m
A = 26,666.67 square meters (approximately)
The dimensions of the rectangle are:
Length = 200 m
Width = 133.33 m
To find the maximum area that can be enclosed in the rectangle, we need to determine the dimensions of the rectangle first.
Let's assume the length of the rectangle is L and the width is W.
It is given that a total of 800 m of fencing material is used to enclose the rectangular field. To calculate the perimeter of the rectangle, we add the lengths of all four sides:
Perimeter = 2L + 3W = 800
Since the field is divided into four portions by fences parallel to one of the sides, the length of the fence parallel to W is divided into three equal parts, resulting in 3W.
We have the equation 2L + 3W = 800.
To find the maximum area, we can express L in terms of W by solving the equation for L:
2L = 800 - 3W
Dividing both sides by 2, we get:
L = (800 - 3W) / 2
Now we can express the area of the rectangle in terms of one variable (W):
Area = L × W = [(800 - 3W) / 2] × W
To maximize the area, we take the derivative of the area function with respect to W and set it equal to zero:
d(Area)/dW = 0
Next, we solve this equation to find the value of W that maximizes the area. Once we find W, we can substitute it back into the equation to find L and calculate the maximum area.
Please note that the above explanation outlines the process to find the maximum area mathematically. However, it may be easier to use a graphing calculator or a computer algebra system to solve the equation and determine the values of W, L, and the maximum area.