Find the maximum area that can be enclosed with a fixed perimeter of fencing

Given a fixed length $L$ of fencing what is the maximum area that can be enclosed in rectangular pasture for which one side is a stone wall (so the length $L$ of fencing need only cover three sides of the rectangle).

Let $y$ denote the width of the pasture (i.e., the length of the sides perpendicular to the wall) and let $x$ denote the length of the pasture (i.e., the length of the side of the rectangle that is parallel to the wall).
Then, we have $2y + x = L$ which implies $x = L - 2y$ . We then want to maximize

$A(y) = (L - 2y)y = Ly - 2y^2.$

Taking the derivative,

$A'(y) = L - 4y.$

Then we have $A'(y) = 0$ when $y = \frac{L}{4}$ and

$\begin{align*} A'(y) &> 0 &\text{when} && y &< \frac{L}{4} \\ A'(y) &< 0 &\text{when} && y &> \frac{L}{4}. \end{align*}$

Therefore, $A(y)$ is increasing when $y < \frac{L}{4}$ and decreasing when $y > \frac{L}{4}$ . Hence, $A(y)$ takes its maximum when $y = \frac{L}{4}$ . Then,