Find the dimensions that maximize the enclosed area.
A renter has 400 feet of fencing to put around the rectangular field and then subdivide the field into three identical smaller rectangular plots by placing two fences parallel to one of the field's shorter sides. Find the dimensions that maximize the enclosed area.
Let L be the length of rectangular field, and W its width.
After the division we got 4 W's (2 for entire field, and 2 for subdivision) and 2 L's, that is:
2L+4W=400 , Let A be the area of the field (A=WL), then
dA/dW = 200-4W
equalize it to zero - to find the extremum; we'll find W=50
So, dimensions that maximize the are for given geometry(including fence's length) would be 100 by 50.