1. **Problem statement:** Find the points where the function $f(x,y) = x + y$ attains its maximum and minimum values subject to the constraint $x^2 + y^2 = 1$. 2. **Method:** Use the method of Lagrange multipliers. We want to maximize/minimize $f(x,y)$ given the constraint $g(x,y) = x^2 + y^2 - 1 = 0$. 3. **Set up Lagrange system:** We introduce a multiplier $\lambda$ and solve: $$\nabla f = \lambda \nabla g$$ which gives: $$\frac{\partial f}{\partial x} = \lambda \frac{\partial g}{\partial x}, \quad \frac{\partial f}{\partial y} = \lambda \frac{\partial g}{\partial y}$$ 4. **Calculate gradients:** $$\nabla f = (1,1), \quad \nabla g = (2x, 2y)$$ 5. **Equations:** $$1 = \lambda 2x \Rightarrow \lambda = \frac{1}{2x}$$ $$1 = \lambda 2y \Rightarrow \lambda = \frac{1}{2y}$$ 6. **Equate lambdas:** $$\frac{1}{2x} = \frac{1}{2y} \Rightarrow x = y$$ 7. **Use constraint:** $$x^2 + y^2 = 1 \Rightarrow 2x^2 = 1 \Rightarrow x^2 = \frac{1}{2} \Rightarrow x = \pm \frac{1}{\sqrt{2}}$$ Since $x = y$, then: $$y = \pm \frac{1}{\sqrt{2}}$$ 8. **Evaluate $f$ at these points:** - At $\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$: $$f = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2}$$ - At $\left(-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right)$: $$f = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = -\sqrt{2}$$ 9. **Conclusion:** - Maximum of $f$ is $\sqrt{2}$ at $\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$. - Minimum of $f$ is $-\sqrt{2}$ at $\left(-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right)$.