= Failure of multiplier stationarity at a singular constraint
The nonzero <gradient> condition in the equality <Lagrange multiplier> rule cannot simply be dropped. Minimize $f(x,y)=x$ subject to $g(x,y)=x^2+y^2=0$. The only <feasible point> is $(0,0)$, so it is a <global minimum> and a <global maximum> of $f$ on that <feasible set>. Nevertheless,
$$
\nabla f(0,0)=(1,0),\qquad\nabla g(0,0)=(0,0),
$$
and no <real number> $\lambda$ satisfies $\nabla f=\lambda\nabla g$. The point is singular for this representation of the <optimization constraint>; the usual <constraint qualification> fails. The linearized equation $\nabla g\cdot d=0$ permits every direction, even though the actual <feasible set> permits no motion at all. This example explains why blindly solving the multiplier equations can miss an optimum.
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