Answer
The largest area that can be enclosed is approximately 4,166,666.67 squared meters.
Work Step by Step
The perimeter of a rectangle is the addition of all its sides; however, there is an extra length, so:
$P=2w+3l$
The area of a rectangle is width times length:
$A=w\cdot l$
We want the area as a function of $l$, so we must re-write $w$ in terms of $l$. Since we know the value of the perimeter, we can solve for $w$:
$10,000=2w+3l$
$10,000-3l=2w+3l-3l$
$10,000-3l=2w$
$(10,000-3l)/2=2w/2$
$w=5000-1.5l$
Now we can substitute $w$ and get the area as a function of its length:
$A(l)=l\cdot (5000-1.5l)$
$A(l)=5000l-1.5l^2$
Now we'll re-write the quadratic equation to vertex form since the vertex is the global maximum:
$A(l)=-1.5(l^2-\frac{10,000}{3}l)$
$A(l)=-1.5(l^2-\frac{10,000}{3}l+(\frac{\frac{10,000}{3}}{2})^2)+(\frac{\frac{10,000}{3}}{2})^2\cdot(1.5)$
$A(l)=-1.5(l^2-\frac{10,000}{3}l+(\frac{10,000}{6})^2)+\frac{10,000^2}{36}\cdot1.5$
$A(l)=-1.5(l-\frac{10,000}{6})^2+\frac{10,000^2}{24}$
So, the maximum area is the y-value of the vertex, which is about 4,166,666.67 squared meters.