Optimization: Critical Points and Lagrange Multipliers

Finding extrema of $f(x,y)$:

1. Compute $\nabla f = (f_x, f_y)$.
2. Solve $\nabla f = \vec{0}$ to find critical points.
3. Classify using the second derivative test.

Second derivative test — At a critical point $(a,b)$, define:

D = f_{xx}f_{yy} - (f_{xy})^2

Lagrange multipliers — To optimise $f(x,y)$ subject to $g(x,y) = 0$:

$$\nabla f = \lambda\,\nabla g \qquad\text{and}\qquad g(x,y) = 0$$

The scalar $\lambda$ is the Lagrange multiplier. It measures how sensitive the optimum is to changes in the constraint.

Geometric meaning: At the optimum, the level curves of $f$ and $g$ are tangent — their gradients are parallel.

Tip: Always check boundary behaviour too. Lagrange multipliers find constrained critical points, but the global optimum might occur at an endpoint.

Common pitfall: $\nabla f = 0$ finds flat points, but in two variables flat includes saddles — max in one direction, min in another. The second-derivative test (via the Hessian determinant) is what separates peaks, pits and passes.

Lagrange Multipliers

To optimise $f(x,y)$ subject to a constraint $g(x,y) = c$, the method of Lagrange multipliers requires:

$$\nabla f = \lambda\,\nabla g$$

At the optimum, the level curve of $f$ is tangent to the constraint curve $g = c$. The multiplier $\lambda$ measures how much the optimum changes if the constraint is relaxed.

This method converts a constrained problem into a system of equations — no substitution or parameterisation needed.