I am trying to perform a Lagrange constraint problem for a simple set of linear equations (I realize this can be solved by substitution) but I'm curious why/how the Lagrange method is failing and I'm getting a conflicting Lagrange multiplier (terminology?). In my example, x and y are constants, S represents the constraint, Psi represents the function.

$$ \Psi(a,b)=(a+b)*x+b*y $$ $$ S(a,b)=a+b=0.5 $$
$$ \nabla\Psi(a,b)=\lambda\nabla S(a,b) $$ $$ \frac{\partial\Psi}{\partial a}=\lambda\frac{\partial S}{\partial a} $$ $$ x =\lambda*1 $$
$$ \frac{\partial\Psi}{\partial b}=\lambda\frac{\partial S}{\partial b} $$ $$ x+y =\lambda*1 $$

  • $\begingroup$ Are there other constraints (non-negative?) on $a$ and $b$? Else, the infimum value of the objective function is $-\infty$. $\endgroup$ – Michael May 27 '15 at 4:12
  • $\begingroup$ Thank you all for the helpful responses. How would the problem change if I added the constraint that 0<a<0.5 and 0<b<0.5? Does this reduce down to checking those end corners (0, 0.5), (0.5, 0)? $\endgroup$ – Carl Johnson May 27 '15 at 14:34
  • $\begingroup$ It depends on what type of Lagrange multiplier theorem you are looking for. For example, you can compute the dual function $d(\lambda) = \inf_{(a,b)\in [0,1/2]^2}[\Psi(a,b) + \lambda (a+b-1/2)]$. Indeed maximizing $d(\lambda)$ over $\lambda \in \mathbb{R}$ gives you the optimal objective function value. $\endgroup$ – Michael May 27 '15 at 15:41
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    $\begingroup$ Or, you can use Kuhn-Tucker gradient conditions, in which case you need 4 additional constraints to ensure $a \geq 0$, $a\leq 1/2$, $b \geq 0$, $b\leq 1/2$. $\endgroup$ – Michael May 27 '15 at 15:42
  • $\begingroup$ Either way, most statements about Lagrange multipliers start by assuming existence of an optimal solution $x^*$, which is not true if the optimal objective value is $-\infty$. $\endgroup$ – Michael May 27 '15 at 15:43

Part of the problem is that your constraint is a line, this isn't a compact set (closed and bounded), so you aren't guaranteed that your function will have an extreme value on the set.

Furthermore your function, being linear, is just going to grow/decay monotonically as you traverse the constraint. So there won't be any critical points for the method to detect.

Another way to think of this,

The gradient of your function is a constant vector, i.e., independent of $a$ and $b$. The gradient of the constraint function is also constant. The method of Lagrange multipliers answers the question "when are these vectors parallel". If the two constant vectors are not parallel then the method cannot give you an answer.


The equivalent problem is to minimize $x/2+by$ over the plane $(x,y)$, which clearly is driven to $-\infty$ when both $x,y \to -\infty$.


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