# What is the Lagrange Multiplier

1. Aug 8, 2016

### Samia qureshi

Can any body explain in simple and easy words "Lagrange Multiplier" What is it? and when it is used? i googled it but that was explained in much difficult words.

2. Aug 8, 2016

### Krylov

The Lagrange multiplier is a additional auxiliary variable that appears when applying Lagrange's technique to solve an optimization problem with equality constraints by converting it to an unconstrained optimization problem. So: You get rid of the constraint at the cost of introducing an extra unknown.

To proceed, I think it is necessary to introduce notation for the function to be maximized (or: minimized) as well as the constraint(s). For this I could recommend the Wikipedia article, which seems quite accessible if you are familiar with some basic multivariable calculus.

3. Aug 8, 2016

### Krylov

Also, since this is a homework thread, it may be useful to post an example of a problem that you are working on in this context, so it can serve as an illustration.

4. Aug 8, 2016

### Samia qureshi

Isn't it similar to Lagrange equation of motions?

5. Aug 8, 2016

### Krylov

Lagrange's formulation of classical mechanics is indeed based on a constrained energy minimization problem (where the constraints are dictated by the system's geometry) and the equations of motion are obtained from the "Lagrangian". The Lagrangian also appears more generally in constrained optimization problems, also in unrelated fields such as economics, but of course there it has another role and interpretation.

Since you asked for an explanation in simple words, I am hesitant to go further, but if you indicate more about the background of your question, perhaps we can give more adequate answers.

6. Aug 8, 2016

### wrobel

I am afraid that a simple explanation does not exist. The so called Lagrange multipliers is a fundamental mathematical construction, it is the reason why this construction arises in very different branches of math.

Consider three vector spaces $X,Y,Z$ these spaces can be infinite dimensional; and two linear operators $B:X\to Y$ and $A:X\to Z$

Assume also that $B(X)=Y$.

Theorem. If $\ker B\subset\ker A$ then there exists a linear operator $\Lambda:Y\to Z$ such that $A=\Lambda B$.

This theorem and its several versions (for Banach spaces and bounded operators etc) is a source of all Lagrage multipliers in different topics.

Last edited: Aug 8, 2016