1
$\begingroup$

On the wikipedia page and indeed in my own opinion the method of Lagrange multipliers as applied to an equality constraint function is as follows,

Extremise $f(x,y)$ subject to $\phi=c$ for some constant $c$. Then we take \begin{equation} F(x,y,\lambda)=f+\lambda(\phi-c)=0 \end{equation} However in my textbook, Mathematical methods for the physical science by M Boas we have the following worked problem,

Extremise $f(x,y)=x^2+y^2$ subject to $y=1-x^2$. Then $F(x,y,\lambda)$ is written, \begin{equation} F(x,y)=x^2+y^2+\lambda(y+x^2) \end{equation}

Shouldn't however we define $F(x,y)$ as, \begin{equation} F(x,y)=x^2+y^2+\lambda (y+x^2-1) \end{equation} My question is therefore, where did the $-1$ go in the example? Is it neglected since it is a constant? Or do I misunderstand the method?

$\endgroup$
  • $\begingroup$ It depends on what you want to do next. If you are just minimizing $F(x,y)$ over all $(x,y)$, then adding $-\lambda$ to that minimization does not change the minimizer $(x^*,y^*)$. Likewise, if you are just finding points $(x,y)$ such that $\nabla F(x,y)=0$, then adding $-\lambda$ does not matter. $\endgroup$ – Michael Jul 8 '15 at 21:54
  • $\begingroup$ Even if it does not matter (as shown in answers), at least for clarity, I would suggest you always do it your way (writing the exact formula of the constraint). $\endgroup$ – Claude Leibovici Jul 9 '15 at 5:15
1
$\begingroup$

The constant does not matter, as only the (total) derivative of $F$ is used in the Lagrange Multipliers method, so additive constants do not affect this. You may include the $c$, or you may not. You could argue that the $F$ found on Wikipedia contains more information about the problem than that in the example, but it proves to be irrelevant to the method. That isn't to say the constant $c$ is irrelevant, merely that it need not be present in the solution at this stage. $c$ is used later on when the condition $\phi=c$ is imposed.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.