Arancher has 400 feet of fencing to put around a rectangular field and then subdivide the field into 22 identical smaller rectangular plots by placing a fence parallel to one of the field's shorter sides. find the dimensions that maximize the enclosed area.

Overall rectangle: 100 ft × 8 16/23 ft ≈ 100 ft × 8.696 ft

Individual pen: 8 16/23 ft × 4 6/11 ft ≈ 8.696 ft × 4.545 ft

Step-by-step explanation:

Half the fence will be used in each of the orthogonal directions to make the pen. That is, the long side of the overall rectangle will be (400 ft/2)/2 = 100 ft. The short side of the overall rectangle will be (400 ft/2)/23 = 8 16/23 ft. (There are 21 partitions between the 22 pens, and 2 end fences, for a total of 23 fence segments of the short length.)

The long (100 ft) side of the overall rectangle is divided into 22 parts by the internal partitions, so each pen will have a short dimension of 100 ft/22 = 4 6/11 ft.

We know half the fence will be used in each direction because we know the total area is a quadratic function of the side length. If the long side of the overall pen is x, then the short side is (400 ft -2x)/23, and the overall area is ...

... A = x(400 ft -2x)/23

The vertex of this quadratic function is halfway between the zeros, at x = 100 ft. That is, the two long sides of the pen total 200 ft, or half the overall length of fence.