The area of chicken farm is x*(20-2x)=-2x? +20x=-2(x-5)？ +50
At this time, when x=5, that is, the length is 10 meter, the chicken farm has the largest area of 50 square meters.
If the side against the wall is wide, the length is (20-x)/2, x≤(20-x)/2, and x≤ 15, that is, 0.
So the area of chicken farm is x*(20-x)/2=-x? /2+ 10x=-(x- 10)？ /2+50
At this time, when x=20/3, that is, the length is 20/3 meters, the area is the largest, 400/9 square meters < 50 square meters.
In order to maximize the area of chicken farm, the length of chicken farm should be 10 meter.
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