Problem 265

Since $f(x) = 2\sin{x}\cos{x} - \sin^{2}{x} + 1$, we can rewrite the function as

$$f(x) = 2\sin{x}\cos{x} + \cos{2x} = \sin{2x} + \cos{2x}.$$

Using the angle sum formula for sine, this becomes

$$f(x) = \sqrt{2}\left(\frac{\sqrt{2}}{2}\sin{2x} + \frac{\sqrt{2}}{2}\cos{2x}\right) = \sqrt{2}\sin\left(2x + \frac{\pi}{4}\right).$$

Given that $f\left(A + \frac{\pi}{8}\right) = \frac{\sqrt{2}}{3}$, we have

$$\sqrt{2}\sin\left(2A + \frac{\pi}{2}\right) = \frac{\sqrt{2}}{3}.$$

Therefore, $\cos{2A} = \frac{1}{3}$. Solving for $\cos^2{A}$ gives us

$$2\cos^2{A} - 1 = \frac{1}{3},$$

which implies that since $A$ is acute ($0 < A < \frac{\pi}{2}$),

$$\cos{A} = \frac{\sqrt{6}}{3} \text{ and } \sin{A} = \sqrt{1 - \cos^2{A}} = \frac{\sqrt{3}}{3}.$$

Given that $a = \sqrt{3}$, by the law of cosines, we have

$$\begin{align*} a^2 &= b^2 + c^2 - 2bc\cos{A} \\ (\sqrt{3})^2 &= b^2 + c^2 - 2bc\cdot\frac{\sqrt{6}}{3}, \end{align*}$$

which implies that

$$b^2 + c^2 \geq 2bc.$$

Therefore,

$$bc \leq \frac{9}{2} + \frac{3\sqrt{6}}{2}.$$

So the area $S$ is

$$S = \frac{1}{2}bc\sin{A} \leq \frac{1}{2}\left(\frac{9}{2} + \frac{3\sqrt{6}}{2}\right)\cdot\frac{\sqrt{3}}{3} = \frac{3(\sqrt{3} + \sqrt{2})}{4}.$$

Hence, the correct choice is:

$$\boxed{A: \frac{3(\sqrt{3} + \sqrt{2})}{4}}.$$