Lagrange multiplier method, proof of

Let $g(x,y)=c$ and $f(x,y)=d$. Taking the derivative of $f$ and $g$ with respect to $t$ gives:

${\displaystyle \frac{\partial f}{\partial t}}={\displaystyle \frac{\partial f}{\partial x}}{x}^{\prime }(t)+{\displaystyle \frac{\partial f}{\partial y}}{y}^{\prime }(t)=0$

and

${\displaystyle \frac{\partial g}{\partial t}}={\displaystyle \frac{\partial g}{\partial x}}{x}^{\prime }(t)+{\displaystyle \frac{\partial g}{\partial y}}{y}^{\prime }(t)=0$

By letting $\overrightarrow{r}=x(t)\widehat{i}+y(t)\widehat{j},$ the partial derivatives can be rewritten as follows:

${\displaystyle \frac{\partial f}{\partial t}}=\mathrm{grad}f\cdot \overrightarrow{r^{\prime }};$  ${\displaystyle \frac{\partial g}{\partial t}}=\mathrm{grad}g\cdot \overrightarrow{r^{\prime }}$

This implies that $\mathrm{grad}f\times \mathrm{grad}g=0,$ thus $\mathrm{grad}f=\lambda \mathrm{grad}g.$ Now this equation can be rewritten as $f_x\widehat{i}+f_y\widehat{j}=\lambda \left(g_x\widehat{i}+g_y\widehat{j}\right).$ Since ${ℝ}^n\mapsto ℝ,$ this equation can be separated into two new equations:

$f_x=\lambda g_x;f_y=\lambda g_y$

Using the above equations, a new function, $F$, can be defined:

$F(x,y,\lambda )=f(x,y)-\lambda g(x,y)$

which can be generalized as:

$F(x,y,\lambda )=f(x,y)-{\displaystyle \sum _{i=1}^m}{\lambda }_i\left[g_i(x,y)\right].$

Title	Lagrange multiplier method, proof of
Canonical name	LagrangeMultiplierMethodProofOf
Date of creation	2013-03-22 15:25:09
Last modified on	2013-03-22 15:25:09
Owner	aplant (12431)
Last modified by	aplant (12431)
Numerical id	6
Author	aplant (12431)
Entry type	Proof
Classification	msc 45C05
Classification	msc 15A42
Classification	msc 15A18