= Lagrange multiplier determinant condition in two variables
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{title2=$f_xg_y-f_yg_x=0$}
At a regular constrained <stationary point> of $f$ on $g=0$, the <Lagrange multiplier> equations $\nabla f=\lambda\nabla g$ make the two <gradients> linearly dependent. Eliminating $\lambda$ gives the displayed zero <determinant>. This condition is necessary, not sufficient for an <extremum>. At a singular constraint point where $\nabla g=0$, the determinant is automatically zero even when no multiplier exists; for example $g=x^2+y^2$ and $f=x$ at the origin.
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