1. **State the problem:** Find the extreme values (maximum and minimum) of the function $$f(x,y) = x^2 y - x^2 - 2 y^2$$ subject to the constraint $$x^2 + y^2 = 1$$ using Lagrange multipliers. 2. **Set up the Lagrange function:** We introduce a multiplier $$\lambda$$ and define $$$L(x,y,\lambda) = f(x,y) - \lambda (g(x,y) - 1) = x^2 y - x^2 - 2 y^2 - \lambda (x^2 + y^2 - 1)$$$ where $$g(x,y) = x^2 + y^2$$ is the constraint function. 3. **Find partial derivatives and set them equal to zero:** $$$\frac{\partial L}{\partial x} = 2 x y - 2 x - 2 \lambda x = 0$$$ $$$\frac{\partial L}{\partial y} = x^2 - 4 y - 2 \lambda y = 0$$$ $$$\frac{\partial L}{\partial \lambda} = -(x^2 + y^2 - 1) = 0$$$ 4. **Rewrite the system:** $$$2 x y - 2 x - 2 \lambda x = 0 \implies 2 x (y - 1 - \lambda) = 0$$$ $$$x^2 - 4 y - 2 \lambda y = 0$$$ $$$x^2 + y^2 = 1$$$ 5. **Analyze cases:** - Case 1: $$x = 0$$ Then from constraint, $$y^2 = 1 \implies y = \pm 1$$. Substitute into second equation: $$0 - 4 y - 2 \lambda y = 0 \implies y(-4 - 2 \lambda) = 0$$ Since $$y \neq 0$$, $$-4 - 2 \lambda = 0 \implies \lambda = -2$$. - Case 2: $$x \neq 0$$ Then $$y - 1 - \lambda = 0 \implies \lambda = y - 1$$. Substitute into second equation: $$x^2 - 4 y - 2 (y - 1) y = 0 \implies x^2 - 4 y - 2 y^2 + 2 y = 0$$ Using constraint $$x^2 = 1 - y^2$$: $$1 - y^2 - 4 y - 2 y^2 + 2 y = 0 \implies 1 - 3 y^2 - 2 y = 0$$ 6. **Solve quadratic in $$y$$:** $$$3 y^2 + 2 y - 1 = 0$$$ Use quadratic formula: $$$y = \frac{-2 \pm \sqrt{4 + 12}}{6} = \frac{-2 \pm 4}{6}$$$ So, $$$y = \frac{2}{6} = \frac{1}{3}$$ or $$y = \frac{-6}{6} = -1$$ 7. **Find corresponding $$x$$ values:** - For $$y = \frac{1}{3}$$: $$$x^2 = 1 - \left(\frac{1}{3}\right)^2 = 1 - \frac{1}{9} = \frac{8}{9}$$$ $$$x = \pm \frac{2 \sqrt{2}}{3}$$ - For $$y = -1$$: $$$x^2 = 1 - 1 = 0 \implies x = 0$$ 8. **Evaluate $$f(x,y)$$ at critical points:** - At $$(0,1)$$: $$$f = 0^2 \cdot 1 - 0^2 - 2 \cdot 1^2 = -2$$$ - At $$(0,-1)$$: $$$f = 0 - 0 - 2 \cdot 1 = -2$$$ - At $$\left(\pm \frac{2 \sqrt{2}}{3}, \frac{1}{3}\right)$$: $$$f = \left(\frac{8}{9}\right) \cdot \frac{1}{3} - \frac{8}{9} - 2 \cdot \left(\frac{1}{3}\right)^2 = \frac{8}{27} - \frac{8}{9} - \frac{2}{9} = \frac{8}{27} - \frac{10}{9} = \frac{8}{27} - \frac{30}{27} = -\frac{22}{27} \approx -0.8148$$$ 9. **Conclusion:** - Maximum value is $$-\frac{22}{27}$$ at $$\left(\pm \frac{2 \sqrt{2}}{3}, \frac{1}{3}\right)$$. - Minimum value is $$-2$$ at $$(0, \pm 1)$$.