Lagrange multiplier method, proof of
ProofOfLagrangeMultiplierMethod2
2005-07-25 17:00:32
2005-07-27 14:50:47
Proof
Lagrange multiplier method
0
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Let $g(x,y)=c$ and $f(x,y)=d$. Taking the derivative of $f$ and $g$ with respect to $t$ gives:
\begin{center}$\displaystyle\frac{\partial f}{\partial t}=\frac{\partial f}{\partial x}x'(t)+\frac{\partial f}{\partial y}y'(t)=0$\end{center}
\begin{center}and\end{center}
\begin{center}$\displaystyle\frac{\partial g}{\partial t}=\frac{\partial g}{\partial x}x'(t)+\frac{\partial g}{\partial y}y'(t)=0$\end{center}
By letting $\vec{r}=x(t)\hat{i}+y(t)\hat{j},$ the partial derivatives can be rewritten as follows:
\begin{center}$\displaystyle\frac{\partial f}{\partial t}=\operatorname{grad} f\cdot \vec{r'};$ \quad $\displaystyle\frac{\partial g}{\partial t}=\operatorname{grad} g\cdot \vec{r'}$\end{center}
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:
\begin{center}$\displaystyle f_x=\lambda g_x;\quad f_y=\lambda g_y$\end{center}
Using the above equations, a new function, $F$, can be defined:
\begin{center}$\displaystyle F(x,y,\lambda)=f(x,y)-\lambda g(x,y)$\end{center}
which can be generalized as:
\begin{center}$\displaystyle F(x,y,\lambda)=f(x,y)-\sum_{i=1}^m \lambda_i\left[g_i(x,y)\right].$\end{center}