Lagrange Multiplier Method: Why is the Langragian function defined as $f(x,y)+\lambda \cdot g(x,y)$?

Edit: As AlexR points out in this comment, there is no mathematical reason behind defining the Lagrangian, except because it makes the Lagrange Multiplier Method easier to memorize. I find this confusing; for me, it is easier to find the critical values of the system of equations.

I still haven't found an explanation why is the Lagrange function defined as:

$$\Lambda(x,y,\lambda) = f(x,y)+\lambda \cdot g(x,y)$$

Every author doesn't explain the procedure and I don't think is necessary to learn Lagrangian Mechanics to understand the formula, some examples:

To incorporate these conditions into one equation, we introduce an auxiliary function

Now, if we're clever we can write a single equation that will capture this idea. This is where the familiar Lagrangian equation comes in:

$$L=f-\lambda(g-c)$$

We can compactly represent both equations at once by writing the Lagrangian:

$$\Lambda(x,\lambda)=f(x)-\lambda g(x)$$

The list goes on, the thing is that "Lagrangian" seems to have different meanings depending on the context: https://en.wikipedia.org/wiki/Lagrangian_%28disambiguation%29

So in this context, What does "Lagrangian function" means, and what are the steps to get to that function?

• The book "Mathematics for Economists" by Carl P. Simon and Lawrence Blume, p. 413 - 417 gives a nice explanation. Commented Aug 10, 2015 at 19:24
• The Lagrangian $L:=f-\lambda g$ is a purely formal device without any intuitive geometrical or physical content. The essential point is the following: In a conditionally stationary point $x$ we necessarily have $\nabla f(x)=\lambda\nabla g(x)$ for some $\lambda\in{\mathbb R}$, whereby this latter equation can be "explained intuitively". Commented Aug 10, 2015 at 19:32
• @NigelOvermars, Nope, it does the same thing: "There is a convenient way of writing this system (6). Form the Lagrangian Function" I really dont understand how the author can summarise a system of three equations with three unknowns into a simple function where the partial derivatives suddenly disappear. Commented Aug 10, 2015 at 20:00

I'm not entirely sure what you're getting at, but I think this maybe helpful to you:

The intuition is that an extreme point of $f(x)$ under the condition $g(x) = 0$ must satisfy $g(x) = 0$ and $\nabla_\nu f(x) = 0$ for any direction $\nu$ tangential to the candidate set $\{g(x) = 0\}$. If this were not the case, we could go a small, positive distance along $\nabla_\nu f$ (or $-\nabla_\nu f$) to improve the function value without missing the constraint. Write these two below each other and get $$\pmatrix{\nabla f + \lambda \nabla g\\g}(x) = 0$$ This however is the $z$-gradient of $f + \lambda g$ if we write $z = (x,\lambda)$.

It turns out that this in fact is the necessary condition for a constrained minimisation (or maximisation).

• That's not quite right: you don't have $\nabla f(x)=0$, you have $\nabla f(x) \cdot u=0$ for every $u$ which is a direction tangent to the surface $g(x)=0$.
– Ian
Commented Aug 10, 2015 at 19:19
• @Ian Yes, that's right. Let me fix this. Commented Aug 10, 2015 at 19:19
• What are the steps to get to the Lagrangian function? Commented Aug 10, 2015 at 19:25
• @JaneSmith It's already shown in the above: the two conditions $\nabla_x f = \lambda \nabla_x g$ (which says "the gradient of $f$ is parallel to the gradient of $g$", hence the gradient of $f$ is orthogonal to the level set of $g$) and $g(x)=0$ can be combined into $\nabla_z(f+\lambda g)=0$, where $z=(x,\lambda)$.
– Ian
Commented Aug 10, 2015 at 19:27
• @AlexR, Thanks for your help. I'm gonna accept your answer and add an edit to my answer to bring attention to this comment. Commented Aug 10, 2015 at 20:26