A rancher has 400 feet of fencing to put around a rectangular field and then subdivide the field into 2 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. Write your answers as fractions reduced to lowest terms.
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Step-by-step explanation:
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
2A+4W2=400W
A=200W-2W2
dA/dW = 200-4W
equalize it to zero - to find the extremum; we'll find W=50
Correspondingly, L=(400-4*50)/2=100
So, dimensions that maximize the are for given geometry(including fence's length) would be 100 by 50.
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