# Lagrange multipliers method question.

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 $$F(x,y,\lambda)=f+\lambda(\phi-c)=0$$ 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, $$F(x,y)=x^2+y^2+\lambda(y+x^2)$$

Shouldn't however we define $F(x,y)$ as, $$F(x,y)=x^2+y^2+\lambda (y+x^2-1)$$ 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?

• 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. – Michael Jul 8 '15 at 21:54
• 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). – Claude Leibovici Jul 9 '15 at 5:15

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.