a farmer wishes to build a fence around a rectangular field and divide it into two equal parts as well.the dividing fence must be parallel to one of the side of the field. you are asked to find dimension of the field of maximum area that you can fence and divide in two equal parts with a wire fence of 1200meter
if the length is x and the width is y, then
2x+3y = 1200
a = xy = x(1200-2x)/3 = 400x - 2x^2/3
max area when x = 400*3/4 = 300
So, the field is 300 by 200
Note that as in all such problems, the fencing is divided equally among the lengths and widths.
30000
To find the dimensions of the field with the maximum area, we need to solve this problem using calculus. Let's assume the length of the field is x meters and the width is y meters.
1. Define the equations:
Let the length of the dividing fence be y.
The total length of the fence is given as 1200 meters, so we have the equation: 2x + 3y = 1200.
The area of the field is given by: A = x * y.
2. Solve for x in terms of y:
Rearrange the first equation to get:
2x = 1200 - 3y
x = (1200 - 3y)/2
3. Substitute x in the area equation:
A = (1200 - 3y)/2 * y
4. Find the derivative of A with respect to y:
dA/dy = (1200 - 3y)/2 - (3y)/2
5. Set the derivative equal to zero and solve for y:
(1200 - 3y)/2 - (3y)/2 = 0
1200 - 3y - 3y = 0
1200 - 6y = 0
6y = 1200
y = 200
6. Substitute y back into the equation for x to find the corresponding value:
x = (1200 - 3(200))/2
x = (1200 - 600)/2
x = 300/2
x = 150
So the dimensions of the field with the maximum area are 150 meters in length and 200 meters in width.
To find the dimensions of the field with maximum area, we can use the concept of calculus and optimization.
Let's assume the length of the field is L and the width is W. Since the fence needs to be parallel to one of the sides, we can divide the field into two equal parts along the length, resulting in two smaller rectangles of dimensions L/2 x W.
The perimeter of the entire fence would be L + 2W + 2(L/2) = 2L + 2W. Since the fence will be used to divide the field into two equal parts, the total length of the fence used for division is L.
Therefore, the total length of the fence required would be: 2L + 2W + L = 3L + 2W.
We know that the total length of the fence is 1200 meters, so we have the equation: 3L + 2W = 1200.
Now, we need to express the area of the field in terms of L and W. The area is given by A = L * W.
To solve for the dimension of the field with maximum area, we need to express the area A in terms of a single variable. In this case, we can express L in terms of W using the equation 3L + 2W = 1200.
Rearranging the equation, we get: L = (1200 - 2W)/3.
Substituting this value of L in the expression for area A, we get: A = W * (1200 - 2W)/3.
To find the maximum value of A, we need to find the critical points by taking the derivative of A with respect to W and setting it equal to zero.
Differentiating A with respect to W, we get: dA/dW = (1200 - 4W)/3.
Setting dA/dW equal to zero, we get: 1200 - 4W = 0.
Solving for W, we find: W = 300.
Substituting this value back into the equation for L, we get: L = (1200 - 2 * 300)/3 = 200.
Therefore, the dimensions of the field with maximum area that can be fenced and divided into two equal parts with a wire fence of 1200 meters are 200 meters by 300 meters.