Answer:
Required dimensions of the rectangle are L = 200 m, W = 100 m
The largest area that can be enclosed is 20,000 sq m.
Step-by-step explanation:
The available length of the fencing = 500 m
Now, Perimeter of a rectangle = SUM OF ALL SIDES = 2(L+B)
But, here once side of the rectangle is NOT FENCED.
So, the required perimeter
= Perimeter of Complete field - Boundary of 1 open side
= 2(L+ W) - L = 2W + L
Now, fencing is given as 500 m
⇒ 2W + L = 500
Now, to maximize the length and width:
put L = 200, W = 100
we get 2(W) +L = 2(200) + 100 = 500 m
Hence, required dimensions of the rectangle are L = 200 m, W = 100 m
The maximized area = Length x Width
= 200 m x 100 m = 20, 000 sq m
Hence, the largest area that can be enclosed is 20,000 sq m.
