1. Set up the rectangle

Since the parabola is $y = 12 - x^2,$ and the rectangle has its base on the $x$-axis with upper two vertices on the parabola, the rectangle must be symmetric about the $y$-axis.

Let the upper right vertex be $(x, y)$ on the parabola. Then: $y = 12 - x^2.$ So the upper left vertex is $(-x, y)$.

Hence:

• width of rectangle $= 2x$
• height of rectangle $= y = 12 - x^2$

Therefore, area is $A(x) = 2x(12 - x^2) = 24x - 2x^3,$ where $0 \le x \le 2\sqrt{3}$.

2. Maximize the area

Differentiate: $A'(x) = 24 - 6x^2.$ For maximum area, set $A'(x)=0.$ Thus, $24 - 6x^2 = 0$ $6x^2 = 24$ $x^2 = 4$ $x = 2$ (we take $x>0$).

Now check second derivative: $A''(x) = -12x.$ At $x=2$, $A''(2) = -24 < 0,$ so this gives a maximum.

3. Compute the maximum area

At $x=2$, $y = 12 - 2^2 = 12 - 4 = 8.$ So the rectangle has:

• width $= 2x = 4$
• height $= 8$

Hence maximum area is $A_{\max} = 4 \cdot 8 = 32.$

4. Match with options

The maximum area is $\boxed{32}.$ So the correct option is D.

5. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

They match.