How to solve constrained optimization problems using Lagrange multipliers?

**Geometric Intuition:** At a constrained extremum, the level curve of $f$ must be tangent to the constraint curve $g(x,y) = 0$. The gradient $\ abla f$ is perpendicular to level curves of $f$, and $\ abla g$ is perpendicular to the constraint. At tangency, these normals are parallel: $\ abla f = \lambda \ abla g$. The scalar $\lambda$ (the Lagrange multiplier) tells you how sensitive the optimal value is to changes in the constraint. **Applied to $f(x,y) = xy$ with $x^2 + y^2 = 1$:** **Step 1:** Compute gradients: $$\ abla f = (y, x), \quad \ abla g = (2x, 2y)$$ **Step 2:** Set $\ abla f = \lambda \ abla g$: \begin{align*} y &= 2\lambda x \\ x &= 2\lambda y \end{align*} **Step 3:** Multiply the first by $x$ and the second by $y$: $$xy = 2\lambda x^2, \quad xy = 2\lambda y^2$$ Thus $2\lambda x^2 = 2\lambda y^2$, so either $\lambda = 0$ or $x^2 = y^2$. If $\lambda = 0$, then $y = 0$ and $x = 0$, which fails $x^2 + y^2 = 1$. So $x^2 = y^2$ which means $y = \pm x$. **Step 4:** Substitute into $x^2 + y^2 = 1$: - If $y = x$: $2x^2 = 1 \Rightarrow x = \pm 1/\sqrt{2}, y = \pm 1/\sqrt{2}$ - If $y = -x$: $2x^2 = 1 \Rightarrow x = \pm 1/\sqrt{2}, y = \mp 1/\sqrt{2}$ **Step 5:** Evaluate $f$: - $f(1/\sqrt{2}, 1/\sqrt{2}) = 1/2$ (maximum) - $f(-1/\sqrt{2}, -1/\sqrt{2}) = 1/2$ (maximum) - $f(1/\sqrt{2}, -1/\sqrt{2}) = -1/2$ (minimum) - $f(-1/\sqrt{2}, 1/\sqrt{2}) = -1/2$ (minimum) To classify max vs min without evaluating $f$, you can use the **bordered Hessian** or simply compare function values. The method extends to multiple constraints: $\ abla f = \sum \lambda_i \ abla g_i$.