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When creating an unconstrained optimization problem from an equality constrained one, the usual way to build the Lagrangian, is by adding a term consisting of a multiplier, multiplied by the equality constraint. Are there problem instances, where it makes better sense to square the equality constraint and then use that in the unconstrained problem?

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  • $\begingroup$ The question is hard to answer without a concret example. Why do you want to square the equality constraint? Moreover, square both sides of an equality constraint introduces ambiguities, for example $x=1$ has only one solution, however, $x^2 =1$ has two solutions. $\endgroup$ – The Pheromone Kid Aug 4 '14 at 13:56
  • $\begingroup$ I heard from someone that this is done for some problems, I was curious to some examples. Will squaring lead to easier problem formulation in those cases? $\endgroup$ – haripkannan Aug 4 '14 at 14:10
  • $\begingroup$ @haripkannan In general it won´t. As Ert has demonstrated, squared constraint $\neq$ constraint. $\endgroup$ – callculus Aug 4 '14 at 16:26
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When the equality constraint has the form $g(x)=0$ then replacing it with $h(x):=g^2(x)=0$ is fatal. Lagrange's method detects the points where $df(x)=\lambda dg(x)$ with a finite $\lambda$, and together with $g(x)=0$ you can expect to obtain a finite number of "conditionally stationary points" on the surface $S: \>g(x)=0$. But $dh(x)=0$ at all points of $S$, so the equation $df(x)=\lambda dh(x)$ will not produce a single point that is of interest.

It is another matter with the objective function $f$. When $f$ appears as a square root (e.g., if $f$ represents a distance between certain points) then no harm is done when you replace $f$ by $f^2$.

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