Diagram is missing.
Adding diagram of the problem:

Solution:
Let , width be x and length be y .
The total amount of fencing is given by
500=5x+2y2y=500−5xy=250−(5/2)x
Maximum area (A)=xy
A=x(250−(5/2)x)A=250x−(5/2)x2
Diff. w.r.t x
dxdA=dxd(250x−25x2)=250−5x
For maximum area:
250−5x=0⇒x=50
y=250−(5/2)x=250−(5/2)(50)=125
Hence, width is 50 ft and length is 125 ft.
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