Answer is: option1

Solution:

1. Let the rectangle have its base on the x-axis from \( (-a, 0) \) to \( (a, 0) \), where \( a > 0 \).
2. The other two vertices will then be at \( (-a, 6 - a^2) \) and \( (a, 6 - a^2) \), since they lie on the parabola \( y = 6 - x^2 \).

The area \( A \) of the rectangle is the product of its width and height:

\( A = \text{width} \times \text{height} = (2a) \times (6 - a^2) = 12a - 2a^3 \)

To find the maximum area, take the derivative of \( A \) with respect to \( a \) and set it to zero:

\( \frac{dA}{da} = 12 - 6a^2 \)

Set the derivative equal to zero:

\( 12 - 6a^2 = 0 \Rightarrow a^2 = 2 \Rightarrow a = \sqrt{2} \)

Substitute \( a = \sqrt{2} \) back into the area formula:

\[ A = 12(\sqrt{2}) - 2(\sqrt{2})^3 = 12\sqrt{2} - 2 \times 2\sqrt{2} = 12\sqrt{2} - 4\sqrt{2} = 8\sqrt{2} \]

The maximum area is \( 8\sqrt{2} \), which corresponds to option (A).