1. **State the problem:** Find the maximum and minimum values of the function $f(x,y) = x^2 + y^2$ subject to the constraint $x^2 + y^2 = 1$ using Lagrange multipliers. 2. **Recall the method of Lagrange multipliers:** To find extrema of $f(x,y)$ subject to a constraint $g(x,y) = 0$, solve the system: $$\nabla f = \lambda \nabla g$$ $$g(x,y) = 0$$ where $\lambda$ is the Lagrange multiplier. 3. **Identify functions:** $$f(x,y) = x^2 + y^2$$ $$g(x,y) = x^2 + y^2 - 1 = 0$$ 4. **Compute gradients:** $$\nabla f = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right) = (2x, 2y)$$ $$\nabla g = (2x, 2y)$$ 5. **Set up equations:** $$2x = \lambda 2x$$ $$2y = \lambda 2y$$ $$x^2 + y^2 = 1$$ 6. **Analyze cases:** - If $x \neq 0$ or $y \neq 0$, then dividing both sides by $2x$ or $2y$ gives $1 = \lambda$. - If $x = 0$ and $y = 0$, this contradicts the constraint $x^2 + y^2 = 1$. 7. **Substitute $\lambda = 1$ back:** The constraint remains $x^2 + y^2 = 1$. 8. **Evaluate $f$ on the constraint:** Since $f(x,y) = x^2 + y^2$, and the constraint is $x^2 + y^2 = 1$, then $$f(x,y) = 1$$ 9. **Interpretation:** The function $f(x,y)$ is constant and equal to 1 on the constraint circle, so the maximum and minimum values of $f$ subject to the constraint are both 1. **Final answer:** $$\max f = 1, \quad \min f = 1$$