Suppose you have 400 meters of fencing and you wish to enclose a rectangular space with two parallel partitions: What is the maximum total area of the enclosed space, in square meters?

If the original rectangle is x m wide and y m long

then the partitions are x m as well
so we have
2y + 4x = 400
y + 2x = 200
y = 200 - 2x

Area = xy
A = x(200-2x) = 200x - 2x^2
dA/dx = 200 - 4x = 0 for a max of A
200 = 4x
x = 50
then y = 200 - 100 = 100

if x = 50 and y = 100, we get a maximum area
which is xy or 50(100) = 5000 m^2

To find the maximum total area of the enclosed space with two parallel partitions using 400 meters of fencing, we can use the concept of optimization.

Let's assume the length of the rectangular space is L and the width is W. Since we have two parallel partitions, the fencing will be used for the perimeter of the rectangle, as well as the two parallel partitions.

Therefore, we can set up the equation for the total fencing length:
2L + 3W = 400

To find the maximum area, we need to express the area in terms of a single variable. Since we have two parallel partitions, the length of the outer rectangle will be L - 2W, and the width will be W.

The area of the outer rectangle can be calculated as A = (L - 2W) * W.

Now, let's solve the equation system step-by-step.

Step 1: Solve the equation 2L + 3W = 400 for L:
2L = 400 - 3W
L = (400 - 3W) / 2

Step 2: Substitute L in the area equation:
A = ((400 - 3W) / 2 - 2W) * W

Step 3: Simplify:
A = (400W - 3W^2 - 4W^2)/2

Step 4: Combine like terms:
A = (400W - 7W^2)/2

Step 5: To find the maximum area, take the derivative of A with respect to W and set it equal to zero:
dA/dW = 400 - 14W = 0

Step 6: Solve for W:
400 - 14W = 0
14W = 400
W = 400/14 โ‰ˆ 28.57

Step 7: Substitute the value of W back into the equation for L:
L = (400 - 3(28.57)) / 2
L = 114.29

Step 8: Calculate the maximum area:
A = (114.29 - 2(28.57)) * 28.57
A โ‰ˆ 228.57 * 28.57
A โ‰ˆ 6531.97 square meters

Therefore, the maximum total area of the enclosed space, with two parallel partitions using 400 meters of fencing, is approximately 6531.97 square meters.

To find the maximum total area of the enclosed space, we need to determine the dimensions of the rectangular space that will result in the largest area.

Let's assume the width of the rectangular space is "x" meters. Since we have two parallel partitions, we'll have three sections of fencing: the upper section, the lower section, and the section between the two partitions.

Since we're enclosing a rectangular space, the upper and lower sections will be equal in length to the width (x). The section between the two partitions will be the remaining length of the fencing, which is 400 - 2x (400 meters minus 2 times the width).

The formula for the area of a rectangle is length multiplied by width. So the total area (A) can be expressed as:

A = (x * x) + (x * (400 - 2x)) + (x * (400 - 2x))

Simplifying this equation:

A = x^2 + x(400 - 2x) + x(400 - 2x)
A = x^2 + 400x - 2x^2 + 400x - 2x^2
A = -4x^2 + 800x

To find the maximum total area, we need to find the value of x that maximizes this quadratic equation.

Since the quadratic equation is in the form -ax^2 + bx, the maximum value occurs at x = -b/2a. In our case, a = -4 and b = 800.

Using the formula x = -b/2a, we find:

x = -(800) / (2 * -4)
x = 800 / 8
x = 100

Hence, the width of the rectangular space that will result in the maximum area is 100 meters.

To find the maximum total area, substitute this width value back into the area equation:

A = -4(100)^2 + 800(100)
A = -4(10000) + 80000
A = -40000 + 80000
A = 40000 square meters

Therefore, the maximum total area of the enclosed space is 40000 square meters.