determine the maximum area of a rectangular field that can be enclosed by 2400 m of fencing

A rectangular field is to be enclosed and divided into 4 equal lots by fences parallel to one of the sides. A total of 10,000 meters of fence are available. Find the area of the largest field that can be enclosed.

To determine the maximum area of a rectangular field, we need to consider the fencing length available.

Let's assume that the rectangular field has sides of length 'l' and 'w'. The perimeter of a rectangle is given by the formula: perimeter = 2(l + w).

Given that we have 2400 m of fencing, we can write the equation as:

2400 = 2(l + w)

Simplifying the equation, we get:

1200 = l + w

Now, we need to express the area of the rectangle, A, in terms of one variable, either l or w. The area of a rectangle is given by the formula: A = l * w.

In this case, it's easier to express the area in terms of one variable, so we'll solve for w:

w = 1200 - l

Substituting this into the area equation, we have:

A = l * (1200 - l)

To find the maximum area, we need to find the maximum value of this quadratic equation. We can do this by finding the vertex of the parabola.

The x-coordinate of the vertex is given by: l = -b / (2a), where a = -1 and b = 1200.

l = -1200 / (2 * -1)
l = 1200 / 2
l = 600

Therefore, the length of the rectangle is 600 m. Substituting this value into the equation for w:

w = 1200 - l
w = 1200 - 600
w = 600

Therefore, the width of the rectangle is also 600 m.

Now we can calculate the maximum area:

A = l * w
A = 600 * 600
A = 360,000 square meters

So, the maximum area of the rectangular field that can be enclosed by 2400 m of fencing is 360,000 square meters.

To determine the maximum area of a rectangular field that can be enclosed by a certain amount of fencing, we need to use the concept of optimization. In this case, we want to find the dimensions of the rectangle that will yield the largest possible area.

Let's assume the length of the rectangular field is L, and the width is W. The perimeter of a rectangle is given by the formula: Perimeter = 2L + 2W. In this case, we have a perimeter of 2400 m, so we can write the equation as: 2L + 2W = 2400.

To maximize the area, we need to express the area of the rectangle in terms of a single variable. The area of a rectangle is given by the formula: Area = L x W. Since we have two variables, L and W, we can use the perimeter equation to express one variable in terms of the other.

Rearranging the perimeter equation, we get: 2L = 2400 - 2W, or L = 1200 - W.

Substituting this value into the area equation, we have: Area = (1200 - W) x W.

To maximize the area, we need to find the maximum value of this quadratic equation. One way to find the maximum value is by using calculus, but in this case, we can simplify the equation:

Area = 1200W - W^2.

This is a quadratic equation in the form f(W) = -W^2 + 1200W, which represents a parabola opening downward. The maximum area occurs at the vertex of the parabola, which is also the axis of symmetry. The axis of symmetry of a parabola given by the equation f(W) = ax^2 + bx + c is given by the expression W = -b/2a.

For our equation f(W) = -W^2 + 1200W, a = -1 (coefficient of W^2) and b = 1200 (coefficient of W). Substituting these values into the axis of symmetry formula, we find W = -1200 / (2 * -1) = 600.

Therefore, the width of the rectangle that yields the maximum area is 600 m. Substituting this value back into the length equation L = 1200 - W, we find L = 1200 - 600 = 600 m as well.

So, the maximum area of the rectangular field that can be enclosed by 2400 m of fencing is 600 m * 600 m = 360,000 square meters.

The maximum area of a rectangel is obtained when the rectangle is a square

so 4sides = 240
1 side = 60
area = 60^2 = 3600 m^2