1. **Problem Statement:** We want to find the largest area of a rectangle with its base on the x-axis and its upper vertices on the parabola defined by $$y = 12 - x^2$$. 2. **Understanding the Setup:** The rectangle is symmetric about the y-axis, so if the base extends from $$-x$$ to $$x$$ on the x-axis, the width is $$2x$$. 3. **Height of the Rectangle:** The height is the y-value of the parabola at $$x$$, which is $$12 - x^2$$. 4. **Area Formula:** The area $$A$$ of the rectangle is width times height: $$ A = 2x(12 - x^2) = 24x - 2x^3 $$ 5. **Finding Critical Points:** To maximize area, find where the derivative $$A'(x)$$ is zero: $$ A'(x) = 24 - 6x^2 $$ Set $$A'(x) = 0$$: $$ 24 - 6x^2 = 0 \\ 6x^2 = 24 \\ x^2 = 4 \\ x = \pm 2 $$ Since $$x$$ represents half the base length and must be non-negative, we take $$x = 2$$. 6. **Check Endpoints:** At $$x=0$$, $$A=0$$; at $$x=\sqrt{12}$$, $$A=0$$. 7. **Calculate Maximum Area:** Substitute $$x=2$$: $$ A = 2 \times 2 \times (12 - 2^2) = 4 \times (12 - 4) = 4 \times 8 = 32 $$ 8. **Dimensions of the Rectangle:** Width = $$2x = 4$$ units, Height = $$12 - x^2 = 8$$ units. **Final answer:** The largest area is $$32$$ square units, with dimensions 4 units by 8 units.