Given that
A farmer has 40 meters of fencing to build a rectangular pen for his prize-winning pigs. If the width of the pen is W meters.
Explanation -
We have to find the greatest area that can be enclosed with this fencing.
The width of the pen is W and the fencing is 40 meters.
So its perimeter is 40 meters.
Let the length of the pen will be L then,
Perimeter = 2 (l + b)
40 = 2(l + W)
40/2 = l + W
l + W = 20
l = 20 - W
So the length is (20 - W) meters.
Now we have to find the area then,
[tex]\begin{gathered} Area=l\times b \\ A=(20-W)\times W \\ Area=20W-W^2 \end{gathered}[/tex]Final answer -
So the maximum area that can be enclosed will be 20W - W^2.