# Lagrange's Multipliers basics

1. Nov 15, 2006

### mtanti

I've just started multi dimensional calculus, among which Langrange's Multipliers. I have some questions which will help me grasp the concepts since I'm a very curious guy...

a) What are you finding exactly with this technique?
b) What is the constraint?
c) What does the extra variable represent?
d) Is there another more intuitive but less effiecient way to obtain the same result?

My lecturer is a very poor teacher and just wants to get on with the syllabus I guess... I would like an explanation on the actual graph of the function and not about the practical usage of it such as in economics...

P.S. I understood the reasoning behind partial differentiation but I'm still shakey on the fact that in 3D planes you can only find a general gradient for each axis and not a single general one.

Thanks!

2. Nov 15, 2006

### HallsofIvy

Staff Emeritus
Here's one way to think about it. Suppose you trying to maximise some function f(x,y,z). Since the gradient always points in the direction of fastest increase, moving in the direction of the gradient vector will move you toward the maximum value. AT the maximum, of course, the gradient vector is 0.

Now suppose you want to maximize f(x,y,z) while staying ON a surface given by g(x,y,z)= constant. You might not be able to move in the direction of grad f but "do the best you can". That is, look at the projection of grad f on the surface and move in that direction. you can keep doing that until there is NO projection: grad f is pointing perpendicular to the surface. Since grad g is also perpendicular to any surface g(x,y,z)= constant, grad f and grad g must be pointing in the same direction: one is a multiple of the other:
$$\nabla f= \lambda \nabla g$$
Of course, $\lambda$ is the "Lagrange multiplier".

I have no idea what you mean by this! What do you mean by "3D planes"? A plane in 3 dimensions? What other kind of plane is there?
And be careful about the word "gradient". Some people use it as a synonym for "derivative" or "slope" but in multi-dimensional calculus, it specifically means the "gradient vector",
$$\nabla f= \frac{\partial f}{\partial x}\vec{i}+ \frac{\partial f}{\partial y}\vec{j}+ \frac{\partial f}{\partial z}\vec{k}$$
But I'm still not clear on what you mean by a "general gradient for each axis".

If you were talking about lines, not planes, I would think you were talking about the "direction angles", the angle a line makes with each coordinate plane in turn and the "direction cosines". Actually, in two dimensions, a line makes two such angles also: the angle at which it crosses the x-axis and the angle at which it crosses the y-axis. But then we have a right triangle: one angle is the complement of the other so cosine of one is the sine of the other. In particular, since sin2+ cos2= 1, if you know one, the other is automatic. In 3 dimensions, it's not that simple but we still have
$$cos^2(\theta_x)+ cos^2(\theta_y)+ cos^2(\theta_z)= 1$$
so that the direction cosines for a line are components of a unit vector in the direction of that line.

3. Nov 15, 2006

### arildno

Let's say we want to find a local critical point for a function f(x,y,z) under the constraint g(x,y,z)=0

Now, consider the function: $F(x,y,z,\lambda)=f+\lambda{g}$
Clearly, the critical points of this function will all lie in the region of (x,y,z) given by the equation g(x,y,z)=0!

Furthermore, WITHIN this region, F is IDENTICAL to f!
Thus, it should be reasonable that f's critical points under the constraint g=0 are identical to F's critical points.

4. Nov 16, 2006

### mtanti

OK, so a contraint need not be intersecting into the function? You are just finding the places where both the constraint's and the function's input variables exist.

Since at g = 0, F is f then both their extrema are identicle. But what about the extrema which are not identicle to f's?

About the shakey part, I was saying that in 2D lines we use f'(x) only to find the general gradient of any point on the graph. But in planes we use 2 general gradients, one for the x axis and one for the y axis, thus partial differentiation. Why can't there be just one function f'(x,y)?

It would be appreciated if focus was place on my other questions as well. Thanks.

5. Nov 30, 2006

### kasparov

lagrange multipliers

I am a research engineer and will really appreciate it if someone could tell me how I could solve the 2D Euler's equation for incompressible fluid using Lagrange Multipliers.

The equations are: