1. **State the problem:** We want to find the maximum value of the function $$y = 2x - x^3$$ and determine the maximum value of $$c$$ such that $$2x - x^3 = c$$ has solutions. 2. **Formula and rules:** To find the maximum of a function, we take its derivative and set it equal to zero to find critical points. 3. **Find the derivative:** $$y' = \frac{d}{dx}(2x - x^3) = 2 - 3x^2$$ 4. **Set derivative to zero to find critical points:** $$2 - 3x^2 = 0$$ $$3x^2 = 2$$ $$x^2 = \frac{2}{3}$$ $$x = \pm \sqrt{\frac{2}{3}} = \pm \frac{\sqrt{6}}{3}$$ 5. **Evaluate the function at the critical points:** At $$x = \frac{\sqrt{6}}{3}$$, $$y = 2x - x^3 = 2 \cdot \frac{\sqrt{6}}{3} - \left(\frac{\sqrt{6}}{3}\right)^3$$ Calculate each term: $$2 \cdot \frac{\sqrt{6}}{3} = \frac{2\sqrt{6}}{3}$$ $$\left(\frac{\sqrt{6}}{3}\right)^3 = \frac{(\sqrt{6})^3}{27} = \frac{6\sqrt{6}}{27} = \frac{2\sqrt{6}}{9}$$ So, $$y = \frac{2\sqrt{6}}{3} - \frac{2\sqrt{6}}{9} = \frac{6\sqrt{6}}{9} - \frac{2\sqrt{6}}{9} = \frac{4\sqrt{6}}{9}$$ 6. **Approximate the maximum value:** $$\frac{4\sqrt{6}}{9} \approx \frac{4 \times 2.449}{9} = \frac{9.796}{9} \approx 1.088$$ 7. **Conclusion:** The maximum value of $$y = 2x - x^3$$ is approximately $$1.088$$, so the constant $$c$$ must be less than or equal to this value for the equation $$2x - x^3 = c$$ to have real solutions. **Note:** The user’s approximation $$1.037$$ is close but slightly different due to a different critical point calculation. The exact critical point is at $$x = \sqrt{\frac{2}{3}}$$, not $$\frac{2}{3}$$. Final answer: $$\max y = \frac{4\sqrt{6}}{9} \approx 1.088$$