Setting up and maximizing the enclosed area

The fence consists of $x + 2y + \frac{\pi y}{2} = 150$. Rearranging: $x + y\left(2 + \frac{\pi}{2}\right) = 150$, so $y = \dfrac{150 - x}{2 + \pi/2} = \dfrac{300 - 2x}{4+\pi}$. The area is $A = xy + \frac{1}{2}\pi\left(\frac{y}{2}\right)^2 = xy + \frac{\pi y^2}{8}$. Substituting the expression for $y$ gives $A(x) = x\cdot\frac{300-2x}{4+\pi} + \frac{\pi}{8}\left(\frac{300-2x}{4+\pi}\right)^2$, valid for $0 < x < 150$.

To find the maximum, we differentiate $A(x)$ with respect to $x$. Since $y(x) = \frac{300-2x}{4+\pi}$ is linear in $x$, both terms of $A$ are quadratic in $x$. The first term gives $\frac{dA_1}{dx} = \frac{300-2x}{4+\pi} + x\cdot\frac{-2}{4+\pi} = \frac{300-4x}{4+\pi}$. The second term gives $\frac{dA_2}{dx} = \frac{\pi}{8}\cdot 2\cdot\frac{300-2x}{4+\pi}\cdot\frac{-2}{4+\pi} = \frac{-\pi(300-2x)}{2(4+\pi)^2}$. Setting $\frac{dA}{dx} = 0$ and solving yields $x = \dfrac{300}{4+\pi}$. Checking the limits: as $x \to 0^+$, $y \to \frac{300}{4+\pi}$ and $A \to 0$; as $x \to 150^-$, $y \to 0$ and $A \to 0$. Thus the critical point is the maximum.