1. **State the problem:** Find the maximum and minimum values of the function $$f(x,y) = x^2 + y$$ subject to the constraint $$g(x,y) = x^2 + y^3 = 1$$ using Lagrange multipliers. 2. **Recall the method:** We introduce a multiplier $$\lambda$$ and solve the system: $$\nabla f = \lambda \nabla g$$ with the constraint $$g(x,y) = 1$$. 3. **Calculate gradients:** $$\nabla f = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right) = (2x, 1)$$ $$\nabla g = \left(\frac{\partial g}{\partial x}, \frac{\partial g}{\partial y}\right) = (2x, 3y^2)$$ 4. **Set up equations:** $$2x = \lambda 2x$$ $$1 = \lambda 3y^2$$ $$x^2 + y^3 = 1$$ 5. **Analyze cases:** - If $$x \neq 0$$, from $$2x = \lambda 2x$$ we get $$\lambda = 1$$. - Substitute $$\lambda = 1$$ into $$1 = 3y^2 \lambda$$ gives $$1 = 3y^2$$ so $$y^2 = \frac{1}{3}$$ and $$y = \pm \frac{1}{\sqrt{3}}$$. 6. **Find corresponding $$x$$ values:** From constraint: $$x^2 + y^3 = 1$$ Calculate $$y^3$$ for $$y = \pm \frac{1}{\sqrt{3}}$$: $$y^3 = y \cdot y^2 = y \cdot \frac{1}{3} = \pm \frac{1}{3\sqrt{3}}$$ So: - For $$y = \frac{1}{\sqrt{3}}$$: $$x^2 = 1 - \frac{1}{3\sqrt{3}}$$ - For $$y = -\frac{1}{\sqrt{3}}$$: $$x^2 = 1 + \frac{1}{3\sqrt{3}}$$ 7. **Calculate function values:** $$f(x,y) = x^2 + y$$ - For $$y = \frac{1}{\sqrt{3}}$$: $$f = \left(1 - \frac{1}{3\sqrt{3}}\right) + \frac{1}{\sqrt{3}} = 1 + \frac{1}{\sqrt{3}} - \frac{1}{3\sqrt{3}} = 1 + \frac{2}{3\sqrt{3}}$$ - For $$y = -\frac{1}{\sqrt{3}}$$: $$f = \left(1 + \frac{1}{3\sqrt{3}}\right) - \frac{1}{\sqrt{3}} = 1 - \frac{2}{3\sqrt{3}}$$ 8. **If $$x=0$$:** From $$2x = \lambda 2x$$, this is true for any $$\lambda$$. From $$1 = 3y^2 \lambda$$, $$\lambda = \frac{1}{3y^2}$$. Constraint: $$0 + y^3 = 1 \implies y^3 = 1 \implies y = 1$$. Calculate $$\lambda$$: $$\lambda = \frac{1}{3(1)^2} = \frac{1}{3}$$. Calculate $$f$$: $$f(0,1) = 0 + 1 = 1$$. 9. **Summary of critical points and values:** - $$\left(\pm \sqrt{1 - \frac{1}{3\sqrt{3}}}, \frac{1}{\sqrt{3}}\right)$$ with $$f = 1 + \frac{2}{3\sqrt{3}}$$ (maximum) - $$\left(\pm \sqrt{1 + \frac{1}{3\sqrt{3}}}, -\frac{1}{\sqrt{3}}\right)$$ with $$f = 1 - \frac{2}{3\sqrt{3}}$$ (minimum) - $$(0,1)$$ with $$f=1$$ (saddle or intermediate value) **Final answer:** Maximum value of $$f$$ is $$1 + \frac{2}{3\sqrt{3}}$$. Minimum value of $$f$$ is $$1 - \frac{2}{3\sqrt{3}}$$.