1. **State the problem:** We want to find the maximum area of a rectangle with its base on the x-axis and its upper vertices on the parabola given by $y = 6 - x^2$. 2. **Set up the formula for the area:** The rectangle's width is $2x$ (from $-x$ to $x$) and its height is $y = 6 - x^2$. So, the area $A$ is: $$A = \text{width} \times \text{height} = 2x \times (6 - x^2)$$ 3. **Write the area function:** $$A(x) = 2x(6 - x^2) = 12x - 2x^3$$ 4. **Find the critical points by differentiating:** $$A'(x) = \frac{d}{dx}(12x - 2x^3) = 12 - 6x^2$$ 5. **Set the derivative equal to zero to find critical points:** $$12 - 6x^2 = 0$$ $$6x^2 = 12$$ $$x^2 = \cancel{\frac{12}{6}}^{2} = 2$$ $$x = \sqrt{2}$$ (we take the positive root since $x$ represents half the width and must be positive) 6. **Calculate the maximum area by substituting $x = \sqrt{2}$ into $A(x)$:** $$A(\sqrt{2}) = 12\sqrt{2} - 2(\sqrt{2})^3 = 12\sqrt{2} - 2(2\sqrt{2}) = 12\sqrt{2} - 4\sqrt{2} = 8\sqrt{2}$$ 7. **Conclusion:** The maximum area of the rectangle is $8\sqrt{2}$. **Answer: (B) 8\sqrt{2}**