Let $S=\{(x,y)\in\mathbb{R}\times\mathbb{R}:x\geq 0,\,y\geq 0,\,y^{2}\leq 4x,\,y^{2}\leq 12-2x$ and $3y+\sqrt{8x}\leq 5\sqrt{8}\}$. If the area of the region $S$ is $a\sqrt{2}$, then $\alpha$ is equal to

Let S={(x,y)∈R×R:x≥ 0, y≥ 0, y²≤ 4x, y²≤ 12-2x and 3y+√8x≤ 5√8}. If the area of the region S is a√2, then α is equal to $x\ge 0, \quad y\ge 0$ $y^{2}\le 4x \Rightarrow y\le 2\sqrt{x}, \quad y^{2}\le 12-2x \Rightarrow y\le \sqrt{12-2x}$ $3y+\sqrt{8}\,x\le 5\sqrt{8}$ $\text{Area}(S)=a\sqrt{2} \Rightarrow a=?$ $(1)\ \tfrac{17}{2}, \quad (2)\ \tfrac{17}{3}, \quad (3)\ \tfrac{17}{4}, \quad (4)\ \tfrac{17}{5}$ $1.\ \text{boundaries: } y=2\sqrt{x},\ y=\sqrt{12-2x},\ \text{line}$ $2.\ \text{which bound is active where}$ $3.\ \text{integrate, match } a\sqrt{2}$ $y=2\sqrt{x} \text{ meets } y=\sqrt{12-2x}: \ 4x=12-2x \Rightarrow x=2$ $P=(2,\ 2\sqrt{2})$ $3y+\sqrt{8}\,x=5\sqrt{8}$ $P:\ 6\sqrt{2}+4\sqrt{2}=10\sqrt{2}=5\sqrt{8}\ \checkmark; \quad y=0 \Rightarrow x=5$ $0\le x\le 2: \text{upper}=2\sqrt{x}$ $2\le x\le 5: \text{upper}=\dfrac{\sqrt{8}}{3}(5-x)=\dfrac{2\sqrt{2}}{3}(5-x)$ $\int_0^2 2\sqrt{x}\,dx=\left[\dfrac{4}{3}x^{3/2}\right]_0^2=\dfrac{4}{3}\cdot 2\sqrt{2}=\dfrac{8\sqrt{2}}{3}$ $\int_2^5 \dfrac{2\sqrt{2}}{3}(5-x)\,dx=\dfrac{2\sqrt{2}}{3}\left[5x-\dfrac{x^2}{2}\right]_2^5$ $=\dfrac{2\sqrt{2}}{3}\cdot\dfrac{9}{2}=3\sqrt{2}$ $\text{Area}=\dfrac{8\sqrt{2}}{3}+3\sqrt{2}=\dfrac{17\sqrt{2}}{3}=a\sqrt{2}$ $\Rightarrow a=\dfrac{17}{3}$ $\boxed{a=\dfrac{17}{3} \ \Rightarrow\ \text{option }(2)}$