Find Area Parameter for Region Bounded by Inequalities

To find the area of the region $S$, we first analyze the given inequalities:

1. $x \ge 0$
2. $y \ge 0$
3. $y^2 \le 4x \implies x \ge \frac{y^2}{4}$
4. $y^2 \le 12 - 2x \implies x \le 6 - \frac{y^2}{2}$
5. $3y + \sqrt{8}x \le 5\sqrt{8} \implies x \le 5 - \frac{3y}{2\sqrt{2}}$

Let us determine the points of intersection for the boundary curves:

• Curve $C_1: x = \frac{y^2}{4}$
• Curve $C_2: x = 6 - \frac{y^2}{2}$
• Line $L: x = 5 - \frac{3y}{2\sqrt{2}}$

Equating $C_1$ and $C_2$: $\frac{y^2}{4} = 6 - \frac{y^2}{2} \implies \frac{3y^2}{4} = 6 \implies y^2 = 8 \implies y = 2\sqrt{2} \quad (\text{since } y \ge 0)$ Substituting $y = 2\sqrt{2}$ into $C_1$, we get $x = 2$. Thus, $C_1$ and $C_2$ intersect at $(2, 2\sqrt{2})$.

Now, check if Line $L$ passes through the point $(2, 2\sqrt{2})$: $3(2\sqrt{2}) + \sqrt{8}(2) = 6\sqrt{2} + 4\sqrt{2} = 10\sqrt{2} = 5\sqrt{8}$ Thus, all three boundary curves intersect at the common point $(2, 2\sqrt{2})$.

Next, we compare the upper bounds for $x$ as functions of $y$, for $y \in [0, 2\sqrt{2}]$: $x_1(y) = 6 - \frac{y^2}{2} \quad \text{and} \quad x_2(y) = 5 - \frac{3y}{2\sqrt{2}}$

Consider the difference $x_1(y) - x_2(y)$: $x_1(y) - x_2(y) = 1 + \frac{3y}{2\sqrt{2}} - \frac{y^2}{2} = -\frac{1}{2} (y - 2\sqrt{2})\left(y + \frac{1}{\sqrt{2}}\right)$

For $y \in [0, 2\sqrt{2}]$, $(y - 2\sqrt{2}) \le 0$ and $\left(y + \frac{1}{\sqrt{2}}\right) > 0$, so $x_1(y) - x_2(y) \ge 0$. This implies $x_2(y) \le x_1(y)$ on the interval $y \in [0, 2\sqrt{2}]$. Hence, $x \le x_2(y)$ is the tighter upper bound.

Moreover, for $y > 2\sqrt{2}$, we have $\frac{y^2}{4} > 6 - \frac{y^2}{2}$, so no point satisfies both $y^2 \le 4x$ and $y^2 \le 12 - 2x$.

Therefore, for $y \in [0, 2\sqrt{2}]$, $x$ ranges from $\frac{y^2}{4}$ to $5 - \frac{3y}{2\sqrt{2}}$.

The area of the region $S$ can be computed by integrating with respect to $y$: $\text{Area} = \int_{0}^{2\sqrt{2}} \left[ \left(5 - \frac{3y}{2\sqrt{2}}\right) - \frac{y^2}{4} \right] dy$

Evaluating the integral step-by-step: $\text{Area} = \left[ 5y - \frac{3y^2}{4\sqrt{2}} - \frac{y^3}{12} \right]_{0}^{2\sqrt{2}}$

Substituting the upper limit $y = 2\sqrt{2}$: $\text{Area} = 5(2\sqrt{2}) - \frac{3(8)}{4\sqrt{2}} - \frac{16\sqrt{2}}{12}$ $\text{Area} = 10\sqrt{2} - 3\sqrt{2} - \frac{4}{3}\sqrt{2}$ $\text{Area} = 7\sqrt{2} - \frac{4}{3}\sqrt{2} = \left(7 - \frac{4}{3}\right)\sqrt{2} = \frac{17}{3}\sqrt{2}$

Given that the area of the region $S$ is $\alpha \sqrt{2}$, we have: $\alpha = \frac{17}{3}$

Hence, the correct option is (B).