<p>If by <span class="math-container">$\lambda_i$</span> you mean the multiplier belonging to constraint (<span class="math-container">$i$</span>), then <span class="math-container">$\lambda_i$</span> and <span class="math-container">$\lambda_j$</span> being positive do mean that these constraints are active/effective/realized as equalities.</p>
<p>Now it is not quite clear to me what you mean by &quot;<em>corner solution</em>&quot;. The constraints (<span class="math-container">$1$</span>),(<span class="math-container">$2$</span>),...,(<span class="math-container">$k$</span>) usually define a polyhedron in <span class="math-container">$\mathbb{R}^n$</span>, where <span class="math-container">$(x_1,x_2,\dots,x_n)$</span> 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 <span class="math-container">$x_i$</span> is zero?</p>
<hr />
<p>If less then <span class="math-container">$n$</span> multipliers are positive, then it is likely your solution is not in a vertex of the polyhedron, but merely on a face of it.</p>
<p>For an example of this, consider
<span class="math-container">$$
U(x_1,x_2,x_3) = x_1x_2x_3,
$$</span>
and the constraints are
<span class="math-container">$$
9 \geq 2x_1 + x_2 + x_3 \tag{1}
$$</span>
<span class="math-container">$$
9 \geq x_1 + 2x_2 + x_3 \tag{2}
$$</span>
and the usual non-negativity constraints.</p>
<p>The optimal solution in this case is <span class="math-container">$(x_1, x_2, x_3) = (2,2,3)$</span>, and the multipliers are <span class="math-container">$\lambda_1 = \lambda_2 = 2 &gt; 0$</span>. The optimal solution is not a vertex of the feasible region, it is a convex combination of the feasible solutions <span class="math-container">$(3,3,0)$</span> and <span class="math-container">$(0,0,9)$</span>. (These two solutions are vertices.)</p>

CC BY-SA 4.0
2022-02-21T19:57:42.037Z
I see that the use of the term corner solution is probably inappropriate here as I'm interested in also seeing if the optimum is defined by one of the vertices from the feasible region of the two constraints. I guess a more appropriate edit of the question would be if  one can immediately jump to the conclusion from inspection of the multipliers values that our optimum lies on a vertex?
EconJohn
CC BY-SA 4.0
2022-02-22T06:54:59.240Z
@EconJohn I gave a 3D example with two positive multipliers.
Giskard
CC BY-SA 4.0
2022-02-22T16:05:13.510Z
Thank you so much for your efforts. Just to confirm, the presence of multiple positive multipliers which indicate multiple binding constraints does not necessarily mean the optimum is a vertex?
EconJohn
CC BY-SA 4.0
2022-02-22T16:40:37.600Z
@EconJohn I have answered this exact question above: "*The optimal solution is not a vertex of the feasible region*"
Giskard
CC BY-SA 4.0
2022-02-21T19:29:49.817Z
2022-02-22T16:41:09.430Z
2022-02-22T16:41:09.430Z
Giskard
<p>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.</p>
<p>To preface suppose we have the following generic utility maximization problem with <span class="math-container">$k$</span> many constraints which hold with equality.</p>
<p><span class="math-container">$$\max U(x_1,...,x_n)$$</span>
subject to
<span class="math-container">$$m_1\geq\sum_{i=1}^nr_i^1 x_i \tag{1}$$</span>
<span class="math-container">$$...$$</span>
<span class="math-container">$$m_k\ge\sum_{i=1}^n r_i^k x_i \tag{k}$$</span></p>
<p>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.</p>
<p>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.</p>
<p>In short if I identify a case where say two multipliers <span class="math-container">$\lambda_i$</span> and <span class="math-container">$\lambda_j$</span> are strictly positive, does that mean the optimum is at a vertex for this problem?</p>

CC BY-SA 4.0
2022-02-21T01:30:07.033Z
A positive Lagrange multiplier means the corresponding constraint holds with equality at the optimum, which you seem to already assume. Or do you also have nonnegativity constraints for the $x$'s and the $\lambda$'s are for those constraints?
Herr K.
CC BY-SA 4.0
2022-02-21T06:57:12.000Z
Perhaps you meant to write $$m_1 \geq \sum_{i=1}^nr_i^1 x_i$$ instead of $$m_1=\sum_{i=1}^nr_i^1 x_i$$?
Giskard
CC BY-SA 4.0
2022-02-21T15:46:38.823Z
@Giskard correct. Edited accordingly
EconJohn
CC BY-SA 4.0
2022-02-21T15:48:15.047Z
@HerrK. The salient assumption is that  $x\in\mathbb{R}_+^l$
EconJohn
CC BY-SA 4.0
2022-02-21T19:30:02.530Z
I added tags to the constraints for easier reference.
Giskard
<microeconomics><mathematical-economics><optimization><lagrangian>
Constrained Optimization with Multiple Constraints: Do multiple strictly positive multipliers imply a solution at a vertex?
EconJohn
# Constrained Optimization with Multiple Constraints: Do multiple strictly positive multipliers imply a solution at a vertex?

(Asked by: username_0 on 2022-02-21)

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?

**Tags:** microeconomics, mathematical-economics, optimization, lagrangian

> **Comments:**
> **username_1**: A positive Lagrange multiplier means the corresponding constraint holds with equality at the optimum, which you seem to already assume. Or do you also have nonnegativity constraints for the $x$'s and the $\lambda$'s are for those constraints?  
> **username_2**: Perhaps you meant to write $$m_1 \geq \sum_{i=1}^nr_i^1 x_i$$ instead of $$m_1=\sum_{i=1}^nr_i^1 x_i$$?  
> **username_0**: @username_2 correct. Edited accordingly  
> **username_0**: @HerrK. The salient assumption is that  $x\in\mathbb{R}_+^l$  
> **username_2**: I added tags to the constraints for easier reference.  

---

## 1 answers

### Answer by username_2 on 2022-02-21. Score: 5 | Accepted answer

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?

---

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.)

> **Comments:**
> **username_0**: I see that the use of the term corner solution is probably inappropriate here as I'm interested in also seeing if the optimum is defined by one of the vertices from the feasible region of the two constraints. I guess a more appropriate edit of the question would be if  one can immediately jump to the conclusion from inspection of the multipliers values that our optimum lies on a vertex?  
> **username_2**: @username_0 I gave a 3D example with two positive multipliers.  
> **username_0**: Thank you so much for your efforts. Just to confirm, the presence of multiple positive multipliers which indicate multiple binding constraints does not necessarily mean the optimum is a vertex?  
> **username_2**: @username_0 I have answered this exact question above: "*The optimal solution is not a vertex of the feasible region*"  

---

