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Lagrange Multipliers Using The Implicit Function Theorem

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Please see the attached file for a properly formatted answer.

There is a bit of confusion when it comes to notation. Theorem 3 and the Inverse function Theorem as stated in the file you provided shared some symbols such as $mathbf{X}, mathbf{U}$ but they are used differently as we point out below. To make things more clear I will use $mathbf{Y}$ instead of $mathbf{U}$ to denote $(x_3,dotsc,x_n)$. I write inside parentheses other comments about notation.

Using the notation in Theorem 3 and its proof given in the attachment (except for $mathbf{Y}$ instead of $mathbf{U}$) let us write the problem in the setting of the Implicit Function Theorem (IFT) as stated in the attached file.

We write $R^n=R^{(n-2)+2}$ (in IFT we use $n$ ``equals'' $n-2$ and $m=2$) and a point
$(x_1,x_2,x_3,dotsc,x_n)=(mathbf{V},mathbf{Y})$, where $mathbf{V}=(x_1,x_2)inR^2$ and
$mathbf{Y}=(x_3,dotsc,x_n)inR^{n-2}$ (we are using ...

Solution Summary

We clarify the use of the Implicit Function Theorem (IFT) in the proof of the theorem of Lagrange multipliers. We define the functions and check the hypotheses to use the IFT. We then check that the consequences provided by the IFT give what we need to prove the theorem of Lagrange multipliers.

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See Also This Related BrainMass Solution

Implicit Function Theorem and Theorem of Lagrange Multipliers

I don't get how they can take the derivative of g_1 and g_2 with respect to x_1 and x_2 when they are defined as
h_1=h_1 (x_3,x_4,...,x_n )=x_1 and h_2=h_2 (x_3,x_4,...,x_n )=x_2

I need a mathematical justification for how this can be written simply as (21) when x_1 and x_2 are defined as functions from other variables: x_3,x_4,...,x_n

Is it by using the chain rule or something else? Please go through the detailed steps for how this can be a normal quadratic matrix as defined in the implicit function theorem and by that showing why this is an applicable form of the implicit function theorem nonsingular matrix in (21):

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