1. Given region

We need the area of

$$S=\{(x,y): x\ge 0,\ y\ge 0,\ y^2\le 4x,\ y^2\le 12-2x,\ 3y+\sqrt 8\,x\le 5\sqrt 8\}.$$

Let us rewrite all constraints in a more usable form.

• From $y^2\le 4x$, $x\ge \frac{y^2}{4}.$
• From $y^2\le 12-2x$, $2x\le 12-y^2 \implies x\le 6-\frac{y^2}{2}.$
• From $3y+\sqrt 8\,x\le 5\sqrt 8$, $x\le \frac{5\sqrt 8-3y}{\sqrt 8}=5-\frac{3y}{\sqrt 8}=5-\frac{3y}{2\sqrt 2}.$

So for a fixed $y\ge 0$, admissible $x$ satisfy

$$\frac{y^2}{4}\le x\le \min\left(6-\frac{y^2}{2},\ 5-\frac{3y}{2\sqrt2}\right).$$

Thus the width of the region at height $y$ is determined by the smaller of the two upper bounds.

---

2. Find where the two upper bounds intersect

Solve

$$6-\frac{y^2}{2}=5-\frac{3y}{2\sqrt2}.$$

This gives

$$1-\frac{y^2}{2}+\frac{3y}{2\sqrt2}=0.$$

Multiply by $2$:

$$2-y^2+\frac{3y}{\sqrt2}=0 \implies y^2-\frac{3y}{\sqrt2}-2=0.$$

Now check $y=2\sqrt2$:

$$(2\sqrt2)^2-\frac{3(2\sqrt2)}{\sqrt2}-2=8-6-2=0.$$

So the relevant nonnegative intersection is

$$y=2\sqrt2.$$

At this value,

$$x=6-\frac{(2\sqrt2)^2}{2}=6-4=2.$$

So the two upper curves meet at $(2,2\sqrt2)$.

---

3. Determine which upper bound is smaller

Take $y=0$:

• parabola bound gives $x\le 6$,
• line gives $x\le 5$.

Hence for small $y$, the line is the active upper bound.

Therefore:

• for $0\le y\le 2\sqrt2$, upper bound is $x=5-\frac{3y}{2\sqrt2};$
• for $y\ge 2\sqrt2$, upper bound is $x=6-\frac{y^2}{2}.$

Now we also need lower bound $x=\frac{y^2}{4}$ to be below the active upper bound.

---

4. Find the maximum possible $y$ in each part

First part:

Require

$$\frac{y^2}{4}\le 5-\frac{3y}{2\sqrt2}.$$

Multiply by $4$:

$$y^2\le 20-3\sqrt2\,y \implies y^2+3\sqrt2\,y-20\le 0.$$

The positive root is

$$y=2\sqrt2$$

(since $8+12-20=0$). So this part indeed runs from $y=0$ to $y=2\sqrt2$.

Second part:

Require

$$\frac{y^2}{4}\le 6-\frac{y^2}{2}.$$

So

$$\frac{3y^2}{4}\le 6 \implies y^2\le 8 \implies 0\le y\le 2\sqrt2.$$

So the second part starts exactly where the first ends and immediately collapses; there is no interval with $y>2\sqrt2$.

Hence the entire region is only between

$$0\le y\le 2\sqrt2,$$

with

$$\frac{y^2}{4}\le x\le 5-\frac{3y}{2\sqrt2}.$$

---

5. Compute the area

Area

$$A=\int_0^{2\sqrt2}\left(5-\frac{3y}{2\sqrt2}-\frac{y^2}{4}\right)dy.$$

Now integrate termwise:

$$A=\left[5y-\frac{3}{2\sqrt2}\cdot \frac{y^2}{2}-\frac{1}{4}\cdot \frac{y^3}{3}\right]_0^{2\sqrt2}.$$

That is,

$$A=\left[5y-\frac{3y^2}{4\sqrt2}-\frac{y^3}{12}\right]_0^{2\sqrt2}.$$

Substitute $y=2\sqrt2$:

• $5y=10\sqrt2$,
• $y^2=8$, so $\frac{3y^2}{4\sqrt2}=\frac{24}{4\sqrt2}=\frac{6}{\sqrt2}=3\sqrt2,$
• $y^3=(2\sqrt2)^3=16\sqrt2$, so $\frac{y^3}{12}=\frac{16\sqrt2}{12}=\frac{4\sqrt2}{3}.$

Therefore,

$$A=10\sqrt2-3\sqrt2-\frac{4\sqrt2}{3} =\left(7-\frac{4}{3}\right)\sqrt2 =\frac{17}{3}\sqrt2.$$

So if the area is $\alpha\sqrt2$, then

$$\alpha=\frac{17}{3}.$$

---

6. Option check

• A: $\frac{17}{2}$ ❌
• B: $\frac{17}{3}$ ✅
• C: $\frac{17}{4}$ ❌
• D: $\frac{17}{5}$ ❌

Thus the correct option is B.