How Do Lagrange Multipliers Extend Beyond Two Variables?

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SUMMARY

Lagrange multipliers are a method used to find extrema of functions subject to constraints, primarily applicable in two-variable scenarios. The discussion highlights that while the method works in R², extending it to R³ and higher dimensions introduces complexities, particularly regarding the relationship between gradients. Specifically, the gradients of the function and the constraint must be parallel, which is not guaranteed in three dimensions if both are perpendicular to a unit vector. The notation "dx/d[Rn]" refers to the directional derivative, which is crucial for understanding the extension of this concept to n-dimensional Euclidean space.

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okkvlt
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How does lagrange multipliers work?
i was able to work out this proof of the idea, but its only true for a function with two independent variables and one dependent variable.

Rn=the space that is the independent variables.

x[Rn]=x
C[Rn]=C=constant.


dx/d[Rn]=grad(x)*v; v is a unit vector
dC/d[Rn]=grad(c)*v

because C is held constant, dC/d[Rn]=0 everywhere.
because cos(pi/2)=0, **grad(C) is perpendicular to v.**

In order for extrema to exist, dx/d[Rn]=0. grad(x)*v is zero meaning **grad(x) is perpendicular to v.**

in the case Rn=R2:
both grad(x) and grad(c) are perpendiular to v, it means grad(x) must be parallel to grad(c).

That is the requirement given by the system
grad(x)=L*grad(c)
C[Rn]=C
where L is the scalar multiplier (upside down y).

but it seems as though this is only true for the R2 case. in 3 dimnensions, if both grad(x) and grad(c) are perpendicular to v, it doesn't necessarily mean grad(x) is parallel to grad(c). It seems like I am missing something.


How do i extend this to more than two independent variables?
 
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What do you mean by "dx/d[Rn]" where Rn is n dimensional Euclidean space? I don't believe that is standard notation.
 
i know it isnt
[Rn]=x,y,z, etc
basically i meant the directional derivative by dx/d[Rn]"
 

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