The maximum area (in sq. units) of a rectangle having its base on the x-axis and its other two vertices on the parabola, $$y = 12 - x^2$$ such that the rectangle lies inside the parabola, is:

Let the rectangle touch the x-axis at the two points $$(-x,0)$$ and $$(x,0)$$. Because the parabola $$y = 12 - x^2$$ is symmetric about the y-axis, choosing the rectangle symmetric as well gives the greatest possible width while still remaining inside the curve.

The height of the rectangle is obtained from the parabola. For any chosen $$x \ge 0$$, the ordinate of the curve is

$$y = 12 - x^2.$$

Hence the four vertices of the rectangle are $$(-x,0),\; (x,0),\; (x,\,12 - x^2),\; (-x,\,12 - x^2).$$

Now we compute its dimensions. The width is the horizontal distance between the two points on the x-axis:

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

The height is simply the y-coordinate of the upper vertices:

$$\text{height} = 12 - x^2.$$

Therefore the area $$A$$ of the rectangle is

$$A(x) = (\text{width})(\text{height}) = 2x\,(12 - x^2) = 24x - 2x^3.$$

To find the maximum area, we differentiate $$A(x)$$ with respect to $$x$$ and set the derivative equal to zero. First, writing the differentiation rule: if $$A(x) = 24x - 2x^3,$$ then

$$\frac{dA}{dx} = \frac{d}{dx}(24x) - \frac{d}{dx}(2x^3).$$

Evaluating each term separately, we use $$\frac{d}{dx}(kx) = k$$ and $$\frac{d}{dx}(x^n) = nx^{n-1}.$$ Hence

$$\frac{d}{dx}(24x) = 24,$$

$$\frac{d}{dx}(2x^3) = 2 \cdot 3x^{2} = 6x^2.$$

So we have

$$\frac{dA}{dx} = 24 - 6x^2.$$

For a maximum or minimum, we set the derivative to zero:

$$24 - 6x^2 = 0.$$

Solving step by step:

$$6x^2 = 24,$$

$$x^2 = 4,$$

$$x = 2 \quad (\text{we take the positive value because } x \ge 0).$$

Substituting $$x = 2$$ back into the formula for the height,

$$y = 12 - (2)^2 = 12 - 4 = 8.$$

Now we compute the corresponding maximum area:

$$A_{\text{max}} = 2x \cdot y = 2 \times 2 \times 8 = 4 \times 8 = 32.$$

To confirm that this critical point is indeed a maximum, we examine the second derivative. Differentiating $$\frac{dA}{dx} = 24 - 6x^2$$ once more, we get

$$\frac{d^2A}{dx^2} = -12x.$$

Evaluating at $$x = 2,$$

$$\frac{d^2A}{dx^2}\Big|_{x=2} = -12 \times 2 = -24 < 0,$$

which is negative, confirming a maximum.

Thus, the largest possible area of the rectangle that fits under the given parabola is

$$32 \text{ square units}.$$

Hence, the correct answer is Option B.