1. Interpret the figure

The parabola is $y=x^2-1.$ Since the rectangle lies below the x-axis, its top side is on the line $y=0$ and its bottom vertices lie on the parabola.

Let the rectangle be symmetric about the $y$-axis, with vertical sides at $x=\pm a,$ where $0\le a\le 1$.

Then:

• Upper vertices on the $x$-axis are $(-a,0)$ and $(a,0)$.
• Lower vertices on the parabola are $(-a,a^2-1)$ and $(a,a^2-1)$.

2. Find dimensions of the rectangle

• Width of rectangle: $2a$
• Height of rectangle: Since the bottom is at $y=a^2-1<0$ and top is at $y=0$, height is $0-(a^2-1)=1-a^2.$

So area is $A(a)=2a(1-a^2)=2a-2a^3.$

3. Maximize the area

Differentiate: $A'(a)=2-6a^2.$

Set $A'(a)=0$: $2-6a^2=0$ $6a^2=2$ $a^2=\frac13$ $a=\frac1{\sqrt3}$ (because $a\ge 0$)

4. Check maximum

$A''(a)=-12a.$ At $a=\frac1{\sqrt3}$, $A''\left(\frac1{\sqrt3}\right)<0,$ so this gives a maximum.

5. Compute maximum area

$A_{\max}=2\left(\frac1{\sqrt3}\right)\left(1-\frac13\right)$ $=\frac{2}{\sqrt3}\cdot \frac23$ $=\frac{4}{3\sqrt3}.$

6. Match with options

Thus the largest area is $\boxed{\frac{4}{3\sqrt3}}$ which is Option C.

7. Comparison with stored answer

Stored correct answer: C

Our derived answer is also C, so they agree.