Lagrange Multipliers: Exploring G, f, and g

In summary, The theorem states that under certain conditions, if f and g are functions from an open part G of R^n to R, both differentiable and with continuous differential, and f(p)=0, Df(p)#0, and g has an extremum on p, then there exists a delta in R such that Dg(p)=delta*Df(p). However, this does not necessarily mean that Dg(p) is equal to 0 if g has an extremum on p. The implicit function theorem guarantees that there is a function y=y(x) that defines an extremum for g on f^-1(0), but this does not necessarily mean that \frac{\partial g}{\partial x}(p)=0=\
  • #1
allistair
20
0
I'm not entirely sure what the english terms are for some of the things I'm about to say but i hope it's clear what I mean exactly. I'n my handbook the theorom is said to be:

Say G is a part (wich is open) of R^n, f and g are functions from G to R (f:G->R, g:G->R) and both are differentiable (and the differential is continues). If f(p)=0, Df(p)#0 and g has an extremum on p (f^-1(0)) then there is a delta (element of R) for which you can write: Dg(p)=delta*Df(p)

But if g has an extremum on p then wouldn't Dg(p) = 0?

thx in advance
 
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  • #2
But if g has an extremum on p then wouldn't Dg(p) = 0?


Not really. Under the given conditions, the implicit function theorem guarantees that the expression [tex]f(p)=f(p_1,p_2)=0[/tex] defines a function [tex]y=y(x), x\in (p_1-\epsilon, p_1+\epsilon)=I[/tex] (or [tex]x=x(y)[/tex], but choose the former for convenience) . So an extremum for g on [tex]f^{-1}(0)[/tex] is in fact the extremum of -say- [tex]g(x,y(x)),x\in I[/tex].

So [tex]\frac{d}{dx}[g(x,y(x))][/tex] is zero at [tex]p[/tex] and not necessarily [tex]\frac{\partial g}{\partial x}(p)=0=\frac{\partial g}{\partial y}(p)[/tex]
 

1. What are Lagrange multipliers and why are they important in science?

Lagrange multipliers are mathematical tools used to optimize a function subject to a set of constraints. They are important in science because they allow us to find the maximum or minimum value of a function while satisfying certain constraints, which is a common problem in many scientific fields.

2. How do Lagrange multipliers work?

Lagrange multipliers work by introducing a new variable, called a multiplier, to the original function and setting up a system of equations that can be solved to find the optimal values for both the original function and the constraints.

3. Can Lagrange multipliers be used for non-linear problems?

Yes, Lagrange multipliers can be used for both linear and non-linear problems. The only requirement is that the constraints can be expressed as equations or inequalities.

4. What is the relationship between Lagrange multipliers and the gradient?

Lagrange multipliers are closely related to the gradient of a function. The gradient of the function and the constraints are used to set up the system of equations that can be solved to find the optimal values. The gradient of the function represents its direction of steepest ascent, while the gradient of the constraints represents the direction of steepest descent.

5. How are Lagrange multipliers used in real-world applications?

Lagrange multipliers have a wide range of applications in science, engineering, economics, and other fields. They can be used to optimize production processes, minimize costs, and maximize profits in economics. In physics, they can be used to find the path of least resistance or the trajectory of a moving object. In chemistry, they can be used to optimize reaction conditions. In general, Lagrange multipliers are used to solve optimization problems in various real-world scenarios.

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