4. A rectangular enclosure is formed by using 800m of fencing. Find the greatest possible area that can be enclosed in this way and the corresponding dimensions of the rectangle.

4 x = 800

x = 200 feet on a side square

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prove that square is max area

A = x y
P = 2 x + 2 y perimeter
so x + y = P/2
A = x (P/2 - x) = -x^2 +P/2 x

x^2 - (P/2) x = A find vertex,complete square
x^2 - (P/2)x + P^2/16 = -A + P^2/16

(x-P/4)^2 = -(A-P^2/16)

vertex at x= P/4 (then y = P/4 too :)
A = (P/4)^2
so square with side = Perimeter/4 is max area

To find the greatest possible area that can be enclosed, we need to determine the dimensions of the rectangle. Let's assume the length of the rectangle is "l" and the width is "w".

Given that the perimeter of the rectangle is 800m, we can equate this to the formula to find the perimeter of a rectangle:

2(l + w) = 800

Let's simplify the equation:

2l + 2w = 800

Now, we can solve this equation for one variable in terms of the other. Let's solve for l:

2l = 800 - 2w
l = 400 - w

Now, we can express the area of the rectangle in terms of one variable:

Area = length × width = l × w = (400 - w) × w

To find the maximum possible area, we need to find the vertex of the quadratic equation. The equation of the area is a parabola that opens downwards, so the vertex will represent the maximum point.

To find the vertex, we can differentiate the area equation with respect to w and set it equal to zero:

d(Area)/dw = 0

-1w + 400 - 2w = 0

-3w + 400 = 0
-3w = -400
w = 400/3

Plugging this value back into the equation for l:

l = 400 - (400/3)
l = (1200 - 400)/3
l = 800/3

Therefore, the greatest possible area is given by:

Area = length × width = (800/3) × (400/3) = 640,000/9 m²

The corresponding dimensions of the rectangle are approximately:
Length = 800/3 m
Width = 400/3 m

To find the greatest possible area that can be enclosed by the rectangular enclosure, we need to maximize the area while using all 800m of fencing. Let's denote the length of the rectangle as L and the width as W.

To start, let's consider the perimeter of the rectangle. The perimeter is the sum of all four sides, which in this case is 800m:

Perimeter = 2L + 2W = 800m

Now we can solve this equation for either L or W in terms of the other variable. Let's solve it for L:

2L = 800m - 2W

L = (800m - 2W) / 2

Next, we need an equation to represent the area of the rectangle. The area of a rectangle is given by length multiplied by width:

Area = L * W

Now substitute the expression for L into the area equation:

Area = [(800m - 2W) / 2] * W

Simplify the equation:

Area = (800mW - 2W^2) / 2

Now we have the area equation in terms of a single variable, W. To find the maximum area, we need to determine the value of W that maximizes this equation. We can achieve this by finding the vertex of the parabola.

The maximum value of the area occurs at the vertex, which can be found using the formula:

W_vertex = -b / (2a)

In this case, a = -2 and b = 800m. Plug these values into the formula:

W_vertex = -800m / (2 * -2)

W_vertex = -800m / -4

W_vertex = 200m

So the width of the rectangle that maximizes the area is 200m.

To find the length, substitute this value of W into the equation for L:

L = (800m - 2(200m)) / 2

L = (800m - 400m) / 2

L = 400m / 2

L = 200m

Therefore, the dimensions of the rectangle that encloses the greatest possible area using 800m of fencing are 200m x 200m, and the corresponding maximum area is 40,000 square meters.