TITLE: The meaning of $\lambda$ in Lagrange Multipliers
QUESTION [12 upvotes]: This is related to two previous questions which I asked about the history of Lagrange Multipliers and intuition behind the gradient giving the direction of steepest ascent.
I am wondering if the constant $\lambda$ in the Lagrange equation $$\nabla f=\lambda \nabla g$$ has any significance. 
For instance, can the sign of $\lambda$ tell us anything about the nature of the critical point? Does its magnitude have any significance?

REPLY [8 votes]: An optimal value of $f(x,y)$ subject to the constraint $g(x,y)=c$ depends on the value of $c$. If you vary $c$, the optimal value of $f$, and its associated $\lambda$ will vary. For a given $c$ let $P(c)$ be the point at which $f$ is optimal, and $f(P(c))$ the optimal value, and $\lambda(c)$ the associated multiplier. How does $f(P(c))$ change as $c$ changes? The chain rule gives us the answer. Assuming that all the functions are differentiable $\frac{df(P(c))}{dc}=\nabla f(P(c)) \cdot P'(c)$ .  By the Lagrange multiplier method we have $\nabla f(P(c))=\lambda(c)\nabla g(P(c))$. Hence  we get
$\frac{df(P(c))}{dc}=\nabla f(P(c)) \cdot P'(c) =\lambda(c) \nabla g(P(c))\cdot P'(c)$. Again by the chain rule, $\nabla g(P(c))\cdot P'(c)=\frac{dg(P(c))}{dc}=\frac{d(c)}{dc}=1$.  And so $\frac{df(P(c))}{dc}=\lambda(c)$ .  That is, as John D. says, the $\lambda(c)$ is the rate at which the optimal value of $f$ varies as  $c$ varies.