# 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:

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

and

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

$\displaystyle\frac{\partial f}{\partial t}=\operatorname{grad}f\cdot\vec{r^{% \prime}};$$\displaystyle\frac{\partial g}{\partial t}=\operatorname{grad}g\cdot\vec{r^{% \prime}}$

This implies that $\operatorname{grad}f\times\operatorname{grad}g=0,$ thus $\operatorname{grad}f=\lambda\operatorname{grad}g.$ Now this equation can be rewritten as $f_{x}\hat{i}+f_{y}\hat{j}=\lambda\left(g_{x}\hat{i}+g_{y}\hat{j}\right).$ Since $\mathbb{R}^{n}\mapsto\mathbb{R},$ this equation can be separated into two new equations:

$\displaystyle f_{x}=\lambda g_{x};\quad f_{y}=\lambda g_{y}$

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

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

which can be generalized as:

$\displaystyle F(x,y,\lambda)=f(x,y)-\sum_{i=1}^{m}\lambda_{i}\left[g_{i}(x,y)% \right].$

Title Lagrange multiplier method, proof of LagrangeMultiplierMethodProofOf 2013-03-22 15:25:09 2013-03-22 15:25:09 aplant (12431) aplant (12431) 6 aplant (12431) Proof msc 45C05 msc 15A42 msc 15A18