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

We have to find the area of the region

$$S=\left\{(x,y)\in\mathbb{R}\times\mathbb{R}:\;x\ge 0,\;y\ge 0,\;y^{2}\le 4x,\;y^{2}\le 12-2x,\;3y+\sqrt 8\,x\le 5\sqrt 8\right\}$$

(1) From $$y^{2}\le 4x$$ we get $$x\ge\dfrac{y^{2}}{4}$$.     This is the region to the right of the right-opening parabola $$y^{2}=4x$$.

(2) From $$y^{2}\le 12-2x$$ we get $$x\le 6-\dfrac{y^{2}}{2}$$.     This is the region to the left of the left-opening parabola $$y^{2}=12-2x$$ whose vertex is at $$(6,0)$$.

(3) From $$3y+\sqrt 8\,x\le 5\sqrt 8$$ write $$\sqrt 8 = 2\sqrt 2$$:

$$3y+2\sqrt 2\,x\le 10\sqrt 2 \;\Longrightarrow\; x\le \dfrac{10\sqrt 2-3y}{2\sqrt 2}=5-\dfrac{3y}{2\sqrt 2}$$

This is the region to the left of the straight line $$3y+2\sqrt 2\,x=10\sqrt 2$$.

Combining (1) and (2):

$$\dfrac{y^{2}}{4}\le x\le 6-\dfrac{y^{2}}{2}$$ For such an $$x$$ to exist we need

$$\dfrac{y^{2}}{4}\le 6-\dfrac{y^{2}}{2} \;\Longrightarrow\; 3y^{2}\le 24 \;\Longrightarrow\; y^{2}\le 8 \;\Longrightarrow\; 0\le y\le 2\sqrt 2$$

Now decide which of the two upper bounds $$x_{1}=6-\dfrac{y^{2}}{2},\qquad x_{2}=5-\dfrac{3y}{2\sqrt 2}$$ is smaller (the region must satisfy both). Define

$$D(y)=x_{2}-x_{1} =\Bigl(5-\dfrac{3y}{2\sqrt 2}\Bigr)-\Bigl(6-\dfrac{y^{2}}{2}\Bigr) =-1-\dfrac{3y}{2\sqrt 2}+\dfrac{y^{2}}{2}$$

$$D(y)=0 \;\Longrightarrow\; \dfrac{y^{2}}{2}-\dfrac{3y}{2\sqrt 2}-1=0 \;\Longrightarrow\; y^{2}-\dfrac{3}{\sqrt 2}y-2=0$$ Solving gives $$y=2\sqrt 2\quad\text{or}\quad y=-\dfrac{1}{\sqrt 2}$$. On $$0\le y\lt 2\sqrt 2$$, we have $$D(y)\lt 0$$, so $$x_{2}

Hence for every $$0\le y\le 2\sqrt 2$$ the $$x$$-limits are

$$x_{\text{min}}=\dfrac{y^{2}}{4},\qquad x_{\text{max}}=5-\dfrac{3y}{2\sqrt 2}$$

The area $$A$$ of the region is therefore

$$A=\int_{0}^{2\sqrt 2} \left[x_{\text{max}}-x_{\text{min}}\right]\,dy =\int_{0}^{2\sqrt 2} \left(5-\dfrac{3y}{2\sqrt 2}-\dfrac{y^{2}}{4}\right)dy$$

Integrate term by term:

$$\int 5\,dy = 5y$$ $$\int -\dfrac{3y}{2\sqrt 2}\,dy = -\dfrac{3}{2\sqrt 2}\cdot\dfrac{y^{2}}{2} = -\dfrac{3y^{2}}{4\sqrt 2}$$ $$\int -\dfrac{y^{2}}{4}\,dy = -\dfrac{1}{4}\cdot\dfrac{y^{3}}{3} = -\dfrac{y^{3}}{12}$$

Thus

$$A=\Bigl[\,5y-\dfrac{3y^{2}}{4\sqrt 2}-\dfrac{y^{3}}{12}\Bigr]_{0}^{2\sqrt 2}$$

At $$y=2\sqrt 2$$:

$$y^{2}=8,\quad y^{3}=16\sqrt 2$$

$$A=5(2\sqrt 2)-\dfrac{3\cdot8}{4\sqrt 2}-\dfrac{16\sqrt 2}{12} =10\sqrt 2-\dfrac{24}{4\sqrt 2}-\dfrac{4\sqrt 2}{3}$$

$$\dfrac{24}{4\sqrt 2}=6/\sqrt 2=3\sqrt 2$$

$$\therefore\;A=10\sqrt 2-3\sqrt 2-\dfrac{4\sqrt 2}{3} =7\sqrt 2-\dfrac{4\sqrt 2}{3} =\dfrac{21\sqrt 2-4\sqrt 2}{3} =\dfrac{17\sqrt 2}{3}$$

Thus $$A=\alpha\sqrt 2$$ with $$\alpha=\dfrac{17}{3}$$.

Option B which is: $$\dfrac{17}{3}$$