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  "body": "\n## Statement\n\nLet $U \\subseteq \\mathbb{R}^n$ be open, let $m \u003c n$, and let $f \\colon U \\to \\mathbb{R}$ and $g = (g_1, \\dots, g_m) \\colon U \\to \\mathbb{R}^m$ be of [[class C^1|0TCBZMEYGN]].\nLet $M = \\{x \\in U : g(x) = 0\\}$, and let $a \\in M$ be a [[local extremum|04CPG9K1C4]] of the restriction of $f$ to $M$.\nSuppose that the [[gradients|0NS2QYVC4Y]] $\\nabla g_1(a), \\dots, \\nabla g_m(a)$ are linearly independent.\nThen there is a unique $\\lambda = (\\lambda_1, \\dots, \\lambda_m) \\in \\mathbb{R}^m$ such that\n$$\\nabla f(a) = \\lambda_1 \\nabla g_1(a) + \\dots + \\lambda_m \\nabla g_m(a).$$\n\n## Remarks\n\nThe numbers $\\lambda_i$ are the Lagrange multipliers. Equivalently, $(a, \\lambda)$ is a critical point of the Lagrangian $L(x, \\lambda) = f(x) - \\lambda \\cdot g(x)$.\nThe hypothesis on the gradients is the constraint qualification. It says that the [[Jacobian matrix|0YF0VXKXGB]] $Dg(a)$ has rank $m$.\nThe proof uses the [[implicit function theorem|0BXBADR94C]]. Near $a$, the set $M$ is the graph of a $C^1$ function of $n - m$ variables. The gradient of $f$ is orthogonal to the tangent space of $M$ at $a$.\nThe condition is necessary, not sufficient. Each candidate point needs a further check, for example a comparison of values on a compact set $M$.\n\n## Explanations\n\n### L1 human\n\nAt a constrained extremum, a small move along the constraint set does not change $f$ to first order.\nSo $\\nabla f(a)$ has no component along the constraint set, and it is a combination of the normals $\\nabla g_i(a)$.\nThis turns a constrained problem into a system of $n + m$ equations in $n + m$ unknowns.\n",
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