# 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:

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

and

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

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

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

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 ${\mathbb{R}}^{n}\mapsto \mathbb{R},$ 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 |