A total of 800m of fencing material is used to enclose a rectangular field and to divide the field into four partitions by fences parallel to the width of the field. Find the dimensions of the largest are of the field that can be fenced.
Let's assume the length of the rectangular field is L and the width is W.
Since there are four partitions parallel to the width, we can divide the field into five equal sections:
fence | partition | fence | partition | fence | partition | fence
Each fence will have a length of W, and there are three fences required. So, the total length of the three fences is 3W.
Similarly, each partition will have a length of L, and there are four partitions required. So, the total length of the four partitions is 4L.
Adding up the lengths of the fences and the partitions, we get the total length of the fencing material used:
3W + 4L = 800m ----- (equation 1)
We need to find the dimensions that maximize the area of the field, which is given by A = L * W.
To proceed, we can solve equation 1 for L:
4L = 800m - 3W
L = (800m - 3W)/4 ----- (equation 2)
Substituting equation 2 into the area formula, we get:
A = [(800m - 3W)/4] * W
A = (800W - 3W^2)/4
To maximize the area, we can take the derivative of A with respect to W and set it equal to zero:
dA/dW = 800/4 - 6W/4 = 0
200 - 1.5W = 0
W = 200/1.5
W = 400/3
Substituting this value of W back into equation 2, we can find the corresponding value of L:
L = (800m - 3(400/3)) / 4
L = (800m - 400m) / 4
L = 400m / 4
L = 100m
Therefore, the dimensions of the largest area of the field that can be fenced are:
Length = 100m and Width = 400/3 m
To find the dimensions of the largest area that can be fenced, we need to maximize the area of the rectangular field.
Let's denote the length of the field as L and the width as W. Since there are four partitions by fences parallel to the width, there will be five segments of fencing material used: two for the length of the field, two for the width of the field, and one for the partition between the four sections.
From the given information, we know that the total length of the material used for the fence is 800m. Since there are two lengths and two widths, the length of the material used for the fence can be expressed as:
2L + 2W + 1W = 800
Simplifying the equation, we have:
2L + 3W = 800
Now, we need to express the area (A) of the field in terms of L:
A = L * W
We want to maximize A while satisfying the fencing constraint. To do this, we can rewrite the equation for the fencing as:
L = (800 - 3W) / 2
Substituting this value of L into the equation for A:
A = (800 - 3W) / 2 * W
Now, we can find the critical points of A by taking the derivative with respect to W and setting it equal to zero:
dA/dW = (800 - 3W) / 2 - (800 - 3W) / 2 = 0
Simplifying, we have:
(800 - 3W) - (800 - 3W) = 0
800 - 3W = 800 - 3W
0 = 0
This means that the derivative is equal to zero for all values of W. Therefore, there are no critical points for A.
Since there are no critical points, we need to consider the endpoints of the feasible region. In this case, W cannot be negative. We also know that W must be less than 800/3, otherwise, the length would be negative, which is not possible.
So, the feasible range for W is 0 < W < 800/3.
To determine the maximum value of A, we can evaluate A at the endpoints and compare the results.
When W = 0, the width is zero, and the area becomes zero as well.
When W = 800/3, the width is at its maximum value, and we can substitute it into the equation for A to find the maximum area.
A = (800 - 3(800/3))/2 * (800/3)
= (800 - 800)/2 * (800/3)
= 0 * (800/3)
= 0
Therefore, when W = 800/3, the area A is equal to zero.
Comparing the areas at the endpoints, we find that the maximum area is zero. This means that the dimensions of the largest possible area of the field that can be fenced is a length of 0m and a width of 0m. However, this does not make practical sense, so we need to reconsider the problem or the given constraints.
It appears that there may be an error in the problem statement or constraints, as it is not possible to enclose a field and create four partitions using 800m of fencing material.