Constrained Optimization with Multiple Constraints: Do multiple strictly positive multipliers imply a solution at a vertex?

Question

This might be a bit of a silly question but I am interested in solving standard economic problems with many constraints and am wondering if there are any shortcuts.

To preface suppose we have the following generic utility maximization problem with $k$ many constraints which hold with equality.

$$\max U(x_1,...,x_n)$$ subject to $$m_1\geq\sum_{i=1}^nr_i^1 x_i \tag{1}$$ $$...$$ $$m_k\ge\sum_{i=1}^n r_i^k x_i \tag{k}$$

The traditional way of solving these sort of problems would be to identify possible optimum considering one constraint at a time and then seeing if it violates any constraints. Its possible however that a corner solution exists where in this case we would seek the values at the vertices on our feasible set as defined by our set of constraints.

This is a tedious problem however I'm wondering if just looking at the values of the Lagrange multipliers associated with each one of these constraints (checking if multiple are positive) to infer if a vertex on our feasible region is indeed the optimum.

In short if I identify a case where say two multipliers $\lambda_i$ and $\lambda_j$ are strictly positive, does that mean the optimum is at a vertex for this problem?

Answer 1

If by $\lambda_i$ you mean the multiplier belonging to constraint ($i$), then $\lambda_i$ and $\lambda_j$ being positive do mean that these constraints are active/effective/realized as equalities.

Now it is not quite clear to me what you mean by "corner solution". The constraints ($1$),($2$),...,($k$) usually define a polyhedron in $\mathbb{R}^n$, where $(x_1,x_2,\dots,x_n)$ is such that all constraints are fulfilled, so this polyhedron is the set/region of feasible solutions. If a solution is in the interior of this set, it is definitely not a corner solution; otherwise I am not sure. E.g., is a solution on the edge (but not in any corner/vertex) of a cube a corner solution? Or perhaps you meant at least one $x_i$ is zero?

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If less then $n$ multipliers are positive, then it is likely your solution is not in a vertex of the polyhedron, but merely on a face of it.

For an example of this, consider $$ U(x_1,x_2,x_3) = x_1x_2x_3, $$ and the constraints are $$ 9 \geq 2x_1 + x_2 + x_3 \tag{1} $$ $$ 9 \geq x_1 + 2x_2 + x_3 \tag{2} $$ and the usual non-negativity constraints.

The optimal solution in this case is $(x_1, x_2, x_3) = (2,2,3)$, and the multipliers are $\lambda_1 = \lambda_2 = 2 > 0$. The optimal solution is not a vertex of the feasible region, it is a convex combination of the feasible solutions $(3,3,0)$ and $(0,0,9)$. (These two solutions are vertices.)