= Solution
For a differentiable objective $f$ and regular constraint $g=r$, a constrained <stationary point> satisfies
$$
\boxed{\nabla f=\lambda\nabla g,\qquad g=r,}
$$
where $\lambda$ is a <Lagrange multiplier>. Solve these equations for both the coordinates and $\lambda$.
Geometrically, the <gradient> $\nabla g$ is normal to the constraint <level set> when $\nabla g\ne0$. Every feasible infinitesimal displacement $\mathbf v$ is tangent to that <level set>, so $\nabla g\cdot\mathbf v=0$. At a constrained <stationary point>, the directional change $\nabla f\cdot\mathbf v$ must also vanish for every such tangent displacement. Thus $\nabla f$ is normal to the same tangent <plane> and is a <scalar multiple> of $\nabla g$. This is the geometric basis of the <Lagrange multiplier> method.
For a second constraint $h=s$, with <linearly independent vectors> $\nabla g$ and $\nabla h$, the feasible tangent directions are perpendicular to both. Their normal space is the <linear span> of these two <gradients>, so introduce a second <Lagrange multiplier>:
$$
\boxed{\nabla f=\lambda\nabla g+\mu\nabla h,\qquad g=r,\quad h=s.}
$$
These equations find candidates; deciding whether they are maxima, minima or <saddle points> requires further analysis. Constraint <points> where the required gradient independence fails must be examined separately.
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