Lagrange Multipliers for two constraints, degenerate case

To optimize $f(x,y,z)$ subject to $g(x,y,z)=h(x,y,z)=0$, we use the Lagrange Multiplier method and solve \begin{equation*} \nabla f=\lambda \nabla g+\mu\nabla h,\quad g=0,\quad h=0. \end{equation*} Geometrically, $\nabla f$ must lie on the normal plane spanned by $\nabla g$ and $\nabla h$. However, it can happen that $\nabla g$ is parallel to $\nabla h$ at certain points, and hence they cannot span the normal plane. In this case, does $\nabla f$ have to be parallel to $\nabla g$ to be a critical point? If yes, how to explain it? If no, can it happen that some critical points are missing? Thanks.

• Should I move this to M.SE or submit a new post there? Apr 20, 2018 at 11:17

Lagrange multipliers with two constraints work well if the gradients $\nabla g$ and $\nabla h$ are linearly independent since this implies that in the intersection of level surfaces for $f$ and $g$ we get a regular curve. Without the requirement of linear independence of $\nabla g$ and $\nabla h$ strange things may happen:

Any closed set $K$ in $\mathbb{R}^n$ is a zero set of a smooth nonnegative function. That is a well known fact. Take a smooth function $\tilde{g}:\mathbb{R}^2\to\mathbb{R}_+$ with the zero set $K$ (so $\nabla\tilde{g}=0$ on $K$) and define $g(x,y,z)=z-\tilde{g}(x,y)$, $h(x,y,z)=z$. Then the constraint $g(x,y,z)=h(x,y,z)=0$ is satisfied by the set $K\times\{0\}\subset\mathbb{R}^3$ only so:

Find max/min of $f(x,y,z)$ subject to the constraint $g(x,y,z)=h(x,y,z)=0$,

is equivalent to

Find max/min of $f(x,y,0)$ on $K$.

Since $K$ can be an arbitrary closed set in $\mathbb{R}^2$, one cannot get any multiplier type conditions. Observe that on the set $K\times \{ 0\}$ the gradients of $g$ and $h$ are equal and hence linearly dependent $\nabla g=\nabla h=\langle 0,0,1\rangle$ on $K$.

• You wrote "Lagrange multipliers with two constraints require the gradients ∇g and ∇h to be linearly independent" That is not true. For instance, if both constraints are linear, KKT is necessary, and Lagrange Multipliers will exist, even if the constraint gradient are not linearly independent. Apr 18, 2018 at 17:51
• @MarkL.Stone OK. You are right. I will modify my answer to make your fully justified complaint go away. Apr 18, 2018 at 18:52

If $\nabla g$ and $\nabla h$ are linearly independent at a given point, then the Linear Independence Constraint Qualification (LICQ) is satisfied, and presuming that $f$, $g$, and $h$ are all continuously differentiable, the Karush–Kuhn–Tucker (KKT) conditions are necessary (must hold) for a local minimum or local maximum.

If $\nabla g$ and $\nabla h$ are not linearly independent, and no other constraint qualification holds, then KKT is not necessary, and need not be satisfied at a local minimum or local maximum (i.e., Lagrange Multipliers satisfying KKT may not exist).