# Which Lagrangian multiplier is correct?

$$\max_{x_1,x_2} f(x_1,x_2) \text{ subject to }g(x_1,x_2)=0.$$ So, for critical points, form Lagrangian$$\mathcal L(x_1,x_2,\lambda)\equiv f(x_1,x_2)-\lambda g(x_1,x_2).$$ and then, $$\dfrac{\partial \mathcal L}{\partial x_1}=\dfrac{\partial f(x_1,x_2)}{\partial x_1}-\lambda\dfrac{\partial g(x_1,x_2)}{\partial x_1}=0$$ $$\dfrac{\partial \mathcal L}{\partial x_2}=\dfrac{\partial f(x_1,x_2)}{\partial x_2}-\lambda\dfrac{\partial g(x_1,x_2)}{\partial x_2}=0$$$$\dfrac{\partial \mathcal L}{\partial \lambda}=g(x_1,x_2)=0.$$ Now let $g(x_1,x_2)=p(x_1,x_2)-b=0$, for some function $p$ and constant $b$. But if we interpret $g$ as $b-p(x_1,x_2)=0$, critical points for $x_1$ and $x_2$ remain the same but $\lambda$ reverses the sign.

So, if the constraint is given as $p(x_1,x_2)=b$ is given, which $\lambda$ is "correct"?(i.e., how should we write constraint?)(Sign of Lagrangian multiplier matters very much in economics.)

• When you reverse the definition of $g$, you are changing the sign of the derivative of $g$ as well. There is no inconsistency. – copper.hat Jan 8 '14 at 6:26
• @copper.hat, thank you. – Silent Jan 8 '14 at 6:29
• I can't understand how Lagrangian multiplier remains same here ,when the derivative sign is changed, doesn't λ sign gets changed? – Silent Jan 8 '14 at 7:03
• If $\lambda$ solves the above problem with $g$, then $-\lambda$ will solve the problem with $g$ replaced by $-g$. – copper.hat Jan 8 '14 at 7:05
• I'm not exactly sure what you mean. In the formula above you have defined the Lagrangian with a negative sign and some particular $g$. At a stationary point $x$, you will have some multiplier value $\lambda$. If you rewrite the problem using $-g$ instead, then at the same stationary point $x$, the multiplier value $-\lambda$ (the value being from the $+g$ solution) will solve the equations. (You must be consistent, just as switching $f$ to $-f$ in an unconstrained problem will change the meaning considerably!) – copper.hat Jan 8 '14 at 7:16