Question:

First, identify the region boundaries in the first quadrant ($x \ge 0, y \ge 0$): 1) $y^2 \le 4x \implies x \ge \dfrac{y^2}{4}$ 2) $y^2 \le 12 - 2x \implies x \le 6 - \dfrac{y^2}{2}$ 3) $3y + 2\sqrt{2}x \le 10\sqrt{2} \implies x \le 5 - \dfrac{3y}{2\sqrt{2}}$ Find the intersection of the two parabolas: $$4x = 12 - 2x \implies 6x = 12 \implies x = 2, \quad y = 2\sqrt{2}$$ The boundary line $3y + 2\sqrt{2}x = 10\sqrt{2}$ also passes through $(2, 2\sqrt{2})$ since $3(2\sqrt{2}) + 2\sqrt{2}(2) = 10\sqrt{2}$. For $y \in [0, 2\sqrt{2}]$, the line lies to the left of the second parabola $x = 6 - y^2/2$, making the line the active upper bound on $x$. Thus, the region is bounded by: $$\dfrac{y^2}{4} \le x \le 5 - \dfrac{3y}{2\sqrt{2}}, \quad \text{for } y \in [0, 2\sqrt{2}]$$ Integrating with respect to $y$ to find the area: $$\text{Area} = \int_{0}^{2\sqrt{2}} \left( 5 - \dfrac{3y}{2\sqrt{2}} - \dfrac{y^2}{4} \right) dy$$ $$= \left[ 5y - \dfrac{3y^2}{4\sqrt{2}} - \dfrac{y^3}{12} \right]_{0}^{2\sqrt{2}}$$ $$= 5(2\sqrt{2}) - \dfrac{3(8)}{4\sqrt{2}} - \dfrac{16\sqrt{2}}{12}$$ $$= 10\sqrt{2} - 3\sqrt{2} - \dfrac{4\sqrt{2}}{3} = 7\sqrt{2} - \dfrac{4\sqrt{2}}{3} = \dfrac{17\sqrt{2}}{3}$$ Comparing this to $\alpha\sqrt{2}$ gives: $$\alpha = \dfrac{17}{3}$$ Thus, the correct option is B.