1. **Problem statement:** We want to find the maximum area of a symmetric trapezoid inscribed under the parabola given by the function $$f(x) = -\frac{1}{2}x^2 + 2$$. 2. **Understanding the trapezoid:** The trapezoid is symmetric about the y-axis, with its lower base on the x-axis between points $(-a,0)$ and $(a,0)$, and its upper base parallel to the x-axis at some height $h$ where $0 < h < 2$. 3. **Key conditions:** - The parabola is $$y = -\frac{1}{2}x^2 + 2$$. - The trapezoid's lower base length is $$2a$$. - The upper base lies on the line $$y = h$$. - The trapezoid's top vertices lie on the parabola, so their x-coordinates satisfy $$f(x) = h$$. 4. **Find the x-coordinates of the top base:** Set $$f(x) = h$$: $$-rac{1}{2}x^2 + 2 = h$$ Rearranged: $$x^2 = 2(2 - h)$$ So the top base endpoints are at $$x = \pm \sqrt{2(2 - h)}$$. 5. **Lengths of bases:** - Lower base length: $$2a$$ (where $$a$$ is the x-intercept of the parabola, i.e., where $$f(x) = 0$$). - Upper base length: $$2\sqrt{2(2 - h)}$$. 6. **Find $$a$$ (x-intercepts of parabola):** Set $$f(x) = 0$$: $$-\frac{1}{2}x^2 + 2 = 0 \Rightarrow x^2 = 4 \Rightarrow a = 2$$. 7. **Height of trapezoid:** The height is simply $$h$$ because the lower base is on the x-axis (y=0) and the upper base is at $$y = h$$. 8. **Area formula for trapezoid:** $$A = \frac{(\text{upper base} + \text{lower base})}{2} \times \text{height}$$ Substitute values: $$A(h) = \frac{2\sqrt{2(2 - h)} + 2a}{2} \times h = \frac{2\sqrt{2(2 - h)} + 4}{2} \times h = (\sqrt{2(2 - h)} + 2)h$$ 9. **Simplify area function:** $$A(h) = h\sqrt{2(2 - h)} + 2h$$ 10. **Maximize area:** We find $$h$$ that maximizes $$A(h)$$ for $$0 < h < 2$$. Calculate derivative: $$A'(h) = \sqrt{2(2 - h)} + h \cdot \frac{d}{dh}\sqrt{2(2 - h)} + 2$$ Compute derivative inside: $$\frac{d}{dh}\sqrt{2(2 - h)} = \frac{1}{2\sqrt{2(2 - h)}} \cdot (-2) = -\frac{1}{\sqrt{2(2 - h)}}$$ So: $$A'(h) = \sqrt{2(2 - h)} + h \left(-\frac{1}{\sqrt{2(2 - h)}}\right) + 2 = \sqrt{2(2 - h)} - \frac{h}{\sqrt{2(2 - h)}} + 2$$ Set $$A'(h) = 0$$: $$\sqrt{2(2 - h)} - \frac{h}{\sqrt{2(2 - h)}} + 2 = 0$$ Multiply both sides by $$\sqrt{2(2 - h)}$$: $$2(2 - h) - h + 2\sqrt{2(2 - h)} = 0$$ Simplify: $$4 - 2h - h + 2\sqrt{2(2 - h)} = 0 \Rightarrow 4 - 3h + 2\sqrt{2(2 - h)} = 0$$ Isolate the root term: $$2\sqrt{2(2 - h)} = 3h - 4$$ Square both sides: $$4 \cdot 2(2 - h) = (3h - 4)^2$$ $$8(2 - h) = 9h^2 - 24h + 16$$ $$16 - 8h = 9h^2 - 24h + 16$$ Bring all terms to one side: $$0 = 9h^2 - 24h + 16 - 16 + 8h = 9h^2 - 16h$$ Factor: $$h(9h - 16) = 0$$ Solutions: $$h = 0$$ (not valid, height must be positive) $$9h - 16 = 0 \Rightarrow h = \frac{16}{9} \approx 1.777...$$ 11. **Calculate maximum area:** Substitute $$h = \frac{16}{9}$$ into $$A(h)$$: $$A\left(\frac{16}{9}\right) = \frac{16}{9} \sqrt{2\left(2 - \frac{16}{9}\right)} + 2 \cdot \frac{16}{9}$$ Calculate inside the root: $$2 - \frac{16}{9} = \frac{18}{9} - \frac{16}{9} = \frac{2}{9}$$ So: $$\sqrt{2 \cdot \frac{2}{9}} = \sqrt{\frac{4}{9}} = \frac{2}{3}$$ Therefore: $$A = \frac{16}{9} \times \frac{2}{3} + \frac{32}{9} = \frac{32}{27} + \frac{32}{9} = \frac{32}{27} + \frac{96}{27} = \frac{128}{27} \approx 4.74$$ **Final answer:** The maximum area of the symmetric trapezoid inscribed under the parabola $$f(x) = -\frac{1}{2}x^2 + 2$$ is $$\boxed{\frac{128}{27} \approx 4.74}$$ square units.