The area (in sq. units) of the largest rectangle $$ABCD$$ whose vertices $$A$$ and $$B$$ lie on the $$x$$-axis and vertices $$C$$ and $$D$$ lie on the parabola, $$y = x^2 - 1$$ below the $$x$$-axis, is:

Let us place the rectangle in a symmetrical way so that the algebra becomes as simple as possible. Because the parabola $$y = x^{2}-1$$ is symmetric about the $$y$$-axis, we may assume that the points are

$$A(-x,\,0), \quad B(x,\,0), \quad D(-x,\,y), \quad C(x,\,y)$$

with $$x \gt 0$$ and with $$y$$ negative (since the rectangle must sit below the $$x$$-axis on that part of the parabola). Points $$C$$ and $$D$$ lie on the parabola, so each must satisfy its equation. Substituting the coordinate of, say, point $$C$$ into the parabola we get

$$y = x^{2} - 1.$$

Notice that for the rectangle to be below the $$x$$-axis we require $$y\lt 0$$. That inequality is automatically met as long as $$x^{2}-1\lt 0,$$ which gives $$x^{2}\lt 1$$ or $$0\lt x\lt 1.$$

Now we compute the dimensions of the rectangle. The base of the rectangle stretches from $$A(-x,0)$$ to $$B(x,0),$$ so the length of the base is

$$\text{base} = x - (-x) = 2x.$$

The height of the rectangle is measured from the $$x$$-axis (where $$y=0$$) downward to $$y$$ (a negative quantity). Hence

$$\text{height} = 0 - y = -\,y.$$

Because $$y = x^{2}-1,$$ this becomes

$$\text{height} = -\bigl(x^{2}-1\bigr) = 1 - x^{2}.$$

So the area $$A(x)$$ of the rectangle, as a function of the variable $$x,$$ is

$$A(x) = (\text{base})(\text{height}) = (2x)\,(1 - x^{2}) = 2x - 2x^{3}.$$

To find the maximum area we differentiate with respect to $$x$$ and set the derivative equal to zero. First, the derivative is

$$\frac{dA}{dx} = \frac{d}{dx}\bigl(2x - 2x^{3}\bigr) = 2 - 6x^{2}.$$

Setting this to zero gives the critical points:

$$2 - 6x^{2} = 0 \;\;\Longrightarrow\;\; 6x^{2} = 2 \;\;\Longrightarrow\;\; x^{2} = \frac{1}{3} \;\;\Longrightarrow\;\; x = \frac{1}{\sqrt{3}}.$$

The value $$x = -1/\sqrt{3}$$ is outside our domain $$x\gt 0$$ for the right half-width, so we ignore it. We now evaluate the area at this positive critical point.

For $$x = \dfrac{1}{\sqrt{3}},$$ we have

$$A\!\Bigl(\tfrac{1}{\sqrt{3}}\Bigr) \;=\; 2\Bigl(\tfrac{1}{\sqrt{3}}\Bigr) - 2\Bigl(\tfrac{1}{\sqrt{3}}\Bigr)^{3}.$$ $$= \frac{2}{\sqrt{3}} - 2\left(\frac{1}{3\sqrt{3}}\right).$$

Simplifying the second term first:

$$2\left(\frac{1}{3\sqrt{3}}\right) = \frac{2}{3\sqrt{3}}.$$

Hence

$$A\!\Bigl(\tfrac{1}{\sqrt{3}}\Bigr) = \frac{2}{\sqrt{3}} - \frac{2}{3\sqrt{3}}.$$

To combine these fractions we factor out the common denominator $$\sqrt{3}$$:

$$A\!\Bigl(\tfrac{1}{\sqrt{3}}\Bigr) = \frac{2}{\sqrt{3}}\left(1 - \frac{1}{3}\right).$$

Inside the parentheses:

$$1 - \frac{1}{3} = \frac{2}{3}.$$

Therefore

$$A\!\Bigl(\tfrac{1}{\sqrt{3}}\Bigr) = \frac{2}{\sqrt{3}} \times \frac{2}{3} = \frac{4}{3\sqrt{3}}.$

Finally, we must ensure this is indeed the largest possible area. Observe that at the endpoints of the interval, namely $$x $$\to$$ 0^{+}$$ and $$x $$\to$$ 1^{-},$$ we have

$$A(0) = 0, \qquad A(1) = 2(1) - 2(1) = 0,$$

both of which are smaller than $$\dfrac{4}{3\sqrt{3}}.$$ Hence the critical point we found does give the maximum.

Hence, the correct answer is Option D.