Let's solve this step by step.

The problem asks for the area of the largest rectangle with its lower base on the $x$-axis and upper vertices on the curve $y = 12 - x^2$.

Step 1: Define the rectangle

1. Assume the rectangle has its base centered at the origin.
2. Let the upper vertices of the rectangle be at $(x, y)$ and $(-x, y)$, where $y = 12 - x^2$.
3. The width of the rectangle is $2x$ (from $-x$ to $x$).
4. The height of the rectangle is $y = 12 - x^2$.

Thus, the area $A$ of the rectangle is: $A(x) = \text{width} \times \text{height} = (2x)(12 - x^2)$

Step 2: Express the area function

Expand $A(x)$: $A(x) = 2x(12 - x^2) = 24x - 2x^3$

Step 3: Maximize the area

To find the maximum area, take the derivative of $A(x)$ with respect to $x$ and set it equal to 0: $A'(x) = \frac{d}{dx}(24x - 2x^3) = 24 - 6x^2$

Set $A'(x) = 0$: $24 - 6x^2 = 0$ $6x^2 = 24 \implies x^2 = 4 \implies x = 2 \text{ or } x = -2$

Since $x$ represents the half-width of the rectangle, we take $x = 2$ (positive value).

Step 4: Calculate the maximum area

Substitute $x = 2$ into the area function $A(x)$: $A(2) = 24(2) - 2(2^3) = 48 - 16 = 32$

Thus, the area of the largest rectangle is 32.

Final Answer:

$\boxed{32 \text{ (D)}}$

Would you like further clarification or detailed steps?

Here are 5 related questions for practice:

1. What is the area of the largest rectangle with vertices on the curve $y = 16 - x^2$?
2. How do you find the maximum area of a rectangle under the curve $y = e^{-x^2}$?
3. Why does taking the derivative and finding critical points help maximize the area?
4. What happens to the rectangle's area as $x$ approaches 0 or very large values?
5. How do you solve this problem if the curve equation is changed to $y = k - ax^2$?

Tip: Always verify the second derivative to ensure the critical point gives a maximum, not a minimum!