math question..............?

if you put as many fields next to each other to share fences as you can, you will be able to fence ten identical fields 400 x 400 m using 24 equal lengths. (The maximum area for minimum perimeter is a square.) Each field is therefore 160,000 sq m.

I view this as a 3 x 3 array of rectangles with an additional, sideways rectangle on one end.
This is done best when the sides, w and l, are proportioned 1:3 w:l or vice versa.

So total perimeter of the sum of the fields:
P = 3w + 3l + 3l + w + w + l
P = 5w + 7l : where 'l' is the larger variable

Area of the fields:
A = l * (3l + w)
A = 3l² + wl

For P = 9600, 3w = l:
9600 = 8.66l
A = 3l² + 1.33l

l = 1107.692 m
A = 3(1107.692)² + 1.33(1107.692)
A = 3,682,423.668 m²

A = 3.682 km²

unanswerable... it depends on whether the fields are adjacent and sharing fencing or separate so they cannot share...

i view this as a 3 x 3 array of rectangles with an additional, sideways rectangle on one end...
this is done proper when the sides, w and l, are proportioned 1:3 w:l or vice versa...

so total perimeter of the sum of the fields:
p = 3w + 3l + 3l + w + w + l
p = 5w + 7l : where 'l' is the larger variable

area of the fields:
a = l * (3l + w)
a = 3l² + wl

for p = 9600, 3w = l:
9600 = 8...66l
a = 3l² + 1...33l

l = 1107...692 m
a = 3(1107...692)² + 1...33(1107...692)
a = 3,682,423...668 m²

a = 3...682 km²

If you assign variables to length and width and optimize the total area with respect to a fixed perimeter, you find that the total fence length in one direction is equal to the total fence length in the orthogonal direction. So, you can postulate a field layout and divide up the fencing according to how many fence segments there are in each direction.

The source link shows four different layouts with the stream segments identified by a dashed blue line. The middle left layout is one proposed in another answer to this question. It has fields that are three times as long as wide. (This shape allows the end field to be the same shape as the others.) There are a total of 44 unit-length fence segments required for this layout, so each field is (9600/44 m) by (3*9600/44 m), about 14.281 hectares.

The top left layout has 10 square fields, as there are a total of 12 fence segments in each direction. Each field is 400 m square, so is 16 hectares. That is probably the intended solution to the problem.

However, if we are allowed some leeway in the course of the stream, we can use the layout in the upper right to achieve even more area. This layout has 9 horizontal fence segments and 15 vertical ones, so each field is (4800/9 m) by (4800/15 m), about 17.067 hectares.
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Let there be h fence segments in the horizontal direction and v fence segments in the vertical direction. Then we can divide the fence so that x is used for horizontal fencing and p-x is used for vertical fencing, where p is the total perimeter fence length. The area for n divisions of the field will be
.. area = n*(x/h)*(p-x)/v
.. = (n/(h*v))*x(p-x) ... where n, h, v, p are all constants of the problem and the layout

This describes a parabola with its vertex at x=p/2, confirming our description of the solution in the first paragraph above.

Blah-blah-blah

Unanswerable. It depends on whether the fields are adjacent and sharing fencing or separate so they cannot share.