Lagrange multiplier with one constraint

Statement

Let $U \subseteq \mathbb{R}^n$ be open with $n \ge 2$, and let $f, g \colon U \to \mathbb{R}$ be of class $C^1$. Let $c \in \mathbb{R}$, and let $a$ be a point of the level set $\{x \in U : g(x) = c\}$ with $\nabla g(a) \ne 0$. If the restriction of $f$ to this level set has a local extremum at $a$, then there is a unique $\lambda \in \mathbb{R}$ with $$\nabla f(a) = \lambda \, \nabla g(a).$$

Remarks

This is the case $m = 1$ of the Lagrange multiplier theorem, applied to $g - c$. Geometrically, the level set of $f$ through $a$ is tangent to the constraint set at $a$, because both gradients are orthogonal to it. Example: to maximise $f(x, y) = x + y$ on $x^2 + y^2 = 1$, solve $(1, 1) = \lambda (2 x, 2 y)$ and $x^2 + y^2 = 1$. The maximum is $\sqrt{2}$, at $(1/\sqrt{2}, 1/\sqrt{2})$. Without the condition $\nabla g(a) \ne 0$ the conclusion can fail. For $g(x, y) = y^2 - x^3$ and $f(x, y) = x$, the minimum on $g = 0$ is at $(0, 0)$, where $\nabla g = 0$ and $\nabla f = (1, 0)$.

Explanations

L1 human

Move along the constraint curve and watch the value of $f$. At the best point, the curve just touches a level curve of $f$, so the two curves have the same normal direction there. The multiplier $\lambda$ is the ratio between the two normal vectors.