The total amount of fencing available is $\mathbf{500 ~feet}$ . This represents the perimeter $P$ of the three fenced sides:
$$
\begin{gathered}
P=2 x+y \\\\
500=2 x+y \quad(\text { Equation } 1) \\\\
y=500-2 x
\end{gathered}
$$
The area $A$ of the rectangular field is:
$$
A=x y \quad \text { (Equation 2) }
$$
$$
\begin{aligned}
& A(x)=x(500-2 x) \\\\
& A(x)=500 x-2 x^2
\end{aligned}
$$
The maximum area occurs at the vertex of the parabola represented by $A(x)$, which we can find by setting the first derivative of the area function to zero.
1. Find the derivative of $A(x)$ with respect to $x$ :
$$
\frac{d A}{d x}=500-4 x
$$
2. Set the derivative to zero and solve for $x$ (to find the critical point):
$$
\begin{gathered}
0=500-4 x \\\\
4 x=500 \\\\
x=\frac{500}{4} \\\\
\mathrm{x}=125 \text { feet }
\end{gathered}
$$
Confirm the maximum :
The second derivative is $\frac{d^2 A}{d x^2}=-4$.
Since the second derivative is negative, the function is concave down, confirming that $x=125$ is a maximum.
$$
\begin{gathered}
y=500-2 x \\\\
y=500-2(125) \\\\
y=500-250 \\\\
\mathbf{y}=\mathbf{2 5 0} \text { feet }
\end{gathered}
$$
The dimensions of the field that will enclose the largest area are $\mathbf{1 2 5}$ feet by $\mathbf{2 5 0}$ feet.
