Question 163, Given the ellipse C: $\frac{x^{2}}{12}+\frac{y^{2}}{4}=1$, find the range of the area of its circumscribed rectangle.


Question 163, Answer: The Monge circle corresponding to the ellipse $\mathrm{C}$ is $\mathrm{C}^{\prime}: x^{2}+\mathrm{y}^{2}=16$. The circumscribed rectangles of the ellipse $\mathrm{C}$ are all inscribed rectangles of $\mathrm{C}^{\prime}$.

On one hand, the largest area among the inscribed rectangles of $C^{\prime}$ is its inscribed square, with an area of 32. When the four vertices of this rectangle are $(0, \pm 4)$ and $(\pm 4,0)$, its area is 32, and it is also a circumscribed rectangle of the ellipse $C$.

On the other hand, the width of the circumscribed rectangles of the ellipse $\mathrm{C}$ is no less than $2 b=4$, and the area of the inscribed rectangles of $C$ with a width of no less than 4 is no less than $16 \sqrt{3}$. When the four vertices of this rectangle are $(\pm 2 \sqrt{3}, \pm 2)$, its area is $16 \sqrt{3}$, and it is also a circumscribed rectangle of the ellipse $C$.
In summary, the range of the area of the circumscribed rectangles of the ellipse $C$ is $[16 \sqrt{3}, 32]$.