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Answer:
The required width of the field that would maximize the area is = 1250 feet
Step-by-step explanation:
Given that:
The total fencing length = 5000 ft
Let consider w to be the width and L to be the length.
Then; the perimeter of the rectangular field by assuming a parallel direction is:
P = 3L + 2w
⇒ 3L + 2w = 5000
3L = 5000 - 2w
Recall that:
The area of the rectangle = L×w
Taking the differentiation of both sides with respect to t; we have:
Then; we set A'(w) to be equal to zero;
So;
5000 = 4w
w = 5000/4
w = 1250
Thus; the required width of the field that would maximize the area is = 1250 feet
Also, the length can now be :
L = (5000 -2500)/3
L = 2500/3 feet
Suppose, the farmer divides the plot parallel to the width; Then 2500/3 feet = 833.33 feet and the length L = 1250 feet.