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When one has a function of more variables $f(x_1,\dotsc,x_n)$ and wants to find its maxima and minima on a subset of $\mathbb{R}^n$ defined by $f_1(x_1,\dotsc,x_n)=c_1,\dotsc,f_k(x_1,\dotsc,x_n)=c_k$ with $k\leq n$, one can use the Lagrange multipliers method. When I took my Analysis 2 course, I was taught I had to define:

$$g(x_1,\dotsc,x_n,\lambda_1,\dotsc,\lambda_k)=f(x_1,\dotsc,x_n)+\sum_{i=1}^k\lambda_k(f_k(x_1,\dotsc,x_n)-c_k),$$

and find maxima and minima for $g$. I have now been helping a friend of mine who is studying economics, and her teacher taught her to put a minus before the sum, not a plus. So what is the right sign, minus or plus? Does it make any difference if I use one instead of the other? I would guess so.

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  • $\begingroup$ The $\lambda$ term can be added or subtracted, it makes no real difference. When I looked at maximising functions like $f(x, y)$ I used a positive sign, when looking at multiple constraints I wrote $\nabla f(p) = \sum_{k=1}^M \lambda_k \nabla g_k (p)$ for some vector $p$. $\endgroup$ – Autolatry Feb 19 '15 at 10:45
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When a condition $F(x_1,\ldots, x_n)=c$ just defines a set, then this set might as well be defined by the condition $G(x_1,\ldots, x_n)=-c$, where $G:=-F$. Using the first form and the Lagrange function $\Phi:=f-\lambda F$ will produce conditionally stationary points and a $\lambda$-value for each of them. Using the same principle with the second form of the condition, i.e., putting $\Phi:=f-\lambda G$ will produce the same conditionally stationary points, but the associated $\lambda$-values now have the opposite sign. In most examples of multivariate analysis the obtained $\lambda$-value is thrown away anyway; it has no geometric meaning.

In economics things are different. Here a surface $F(x_1,\ldots, x_n)=c$ separates states where some good costs less than $c$ from the states where it costs more than $c$. It follows that the direction of $\nabla F$ carries economical information which should not be thrown away. Therefore it pays to take care of the sign of $\lambda$. I'm not a mathematical economist, whence I cannot tell you what the exact conventions are.

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There is no 'right sign', you will just find values for $\lambda$ with opposite signs, but since you're interested in the critical points (to find the min/max), the sign of $\lambda$ isn't relevant. It won't influence the critical points you'll find when solving the system.

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I ran into this problem too. My mistake was that I was minimizing $f=h + \lambda g$ but I calculated the multiplier as $\nabla f=\lambda\nabla g$. For that problem formulation, the correct multiplier is $h=-\lambda g$.

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