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\textbf{Goal.} Minimize $f(x)$ while staying on the surface $C(x)=0$.
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\uncover<2->{\textbf{Feasible set as a surface.} Think of $C(x)=0$ as a smooth surface embedded in $\mathbb{R}^n$ (a manifold).}
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\uncover<3->{\textbf{Move without breaking the constraint.} Tangent directions are the “along-the-surface” moves that keep $C(x)$ unchanged to first order. Intuitively: tiny steps that slide on the surface.}
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\uncover<4->{\textbf{What must be true at the best point.} At $x^\star$, there is no downhill direction that stays on the surface. Equivalently, the usual gradient of $f$ has \emph{no component along the surface}.}
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\uncover<5->{\textbf{Normals enter the story.} If the gradient can’t point along the surface, it must point \emph{through} it—i.e., it aligns with a combination of the surface’s normal directions (one normal per constraint).}
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\end{frame}
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% ==== Slide 2: From picture to KKT ====
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\begin{frame}[t]{From the picture to KKT (equality case)}
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\setbeamercovered{invisible}
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\textbf{KKT conditions at a regular local minimum (equality only):}
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\uncover<1->{\textbf{1) Feasibility:} $C(x^\star)=0$. \emph{(We’re on the surface.)}}
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\uncover<2->{\textbf{2) Stationarity:} $\nabla f(x^\star) + J_C(x^\star)^{\!T}\lambda^\star = 0$. \emph{(The gradient is a linear combination of the constraint normals.)}}
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\uncover<3->{\textbf{Lagrangian viewpoint.} Define $L(x,\lambda)=f(x)+\lambda^{\!T}C(x)$. At a solution, $x^\star$ is a stationary point of $L$ w.r.t.\ $x$ (that’s the stationarity equation), while $C(x^\star)=0$ enforces feasibility.}
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\uncover<4->{\textbf{What the multipliers mean.} The vector $\lambda^\star$ tells how strongly each constraint “pushes back” at the optimum; it also measures sensitivity of the optimal value to small changes in the constraints.}
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\end{frame}
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\begin{frame}{KKT system for equalities (first-order necessary conditions)}
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