Lagrange multiplier

Say we have a function $f:\R^n \to \R$ constrainted by an equation $g(\ldots) = c$ where $g: \R^n \to \R$. We can find the extrema of $f$ by finding the points where the $\nabla f$ is parallel to $\nabla g$. Formally, the extrema of $f$ are the solutions to $\nabla f = \lambda \nabla g$ where $\lambda$ is the Lagrange multiplier.

The specific value for $\lambda$ does not matter, it only matters that some $\lambda$ exists for which the equation holds.

Consider why this works. $\nabla g$ is always perpendicular to the level curve $g=c$. If $\nabla f$ is not parallel to $\nabla g$, then we can follow $\nabla f$ along the level curve $g=c$ to reach a different $f$. But when $\nabla f \parallel \nabla g$, we have nowhere to move on $g=c$ that will change $f$, so we have found a local extremum.