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  "body": "\n## Statement\n\nLet $U \\subseteq \\mathbb{R}^n$ be open with $n \\ge 2$, and let $f, g \\colon U \\to \\mathbb{R}$ be of class $C^1$.\nLet $c \\in \\mathbb{R}$, and let $a$ be a point of the [[level set|0Y5Q8Q9WJ7]] $\\{x \\in U : g(x) = c\\}$ with $\\nabla g(a) \\ne 0$.\nIf the restriction of $f$ to this level set has a local extremum at $a$, then there is a unique $\\lambda \\in \\mathbb{R}$ with\n$$\\nabla f(a) = \\lambda \\, \\nabla g(a).$$\n\n## Remarks\n\nThis is the case $m = 1$ of the [[Lagrange multiplier theorem|0VKKP8X6Q6]], applied to $g - c$.\nGeometrically, the level set of $f$ through $a$ is tangent to the constraint set at $a$, because both [[gradients are orthogonal|0VXYV3NQQ1]] to it.\nExample: 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 $(1/\\sqrt{2}, 1/\\sqrt{2})$.\nWithout 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 $\\nabla f = (1, 0)$.\n\n## Explanations\n\n### L1 human\n\nMove along the constraint curve and watch the value of $f$.\nAt the best point, the curve just touches a level curve of $f$, so the two curves have the same normal direction there.\nThe multiplier $\\lambda$ is the ratio between the two normal vectors.\n",
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