# Lagrange multiplier method

The Lagrange multiplier method is used when one needs to find the extreme or stationary points of a function on a set which is a subset of the domain.

Method

Suppose that $f(\mathbf{x})$ and ${g}_{i}(\mathbf{x}),i=1,\mathrm{\dots},m$ ($\mathbf{x}\in {\mathbb{R}}^{n}$) are differentiable functions that map ${\mathbb{R}}^{n}\mapsto \mathbb{R}$, and we want to solve

$$\mathrm{min}f(\mathbf{x}),\mathrm{max}f(\mathbf{x})\mathit{\hspace{1em}}\text{such that}\mathit{\hspace{1em}}{g}_{i}(\mathbf{x})=0,i=1,\mathrm{\dots},m$$ |

By a calculus theorem^{}, if the constaints are independent, the gradient^{} of $f$, $\nabla f$, must satisfy the following equation at the stationary points:

$$\nabla f=\sum _{i=1}^{m}{\lambda}_{i}\nabla {g}_{i}$$ |

The constraints are said to be independent iff all the gradients of each constraint are linearly independent^{}, that is:

$\{\nabla {g}_{1}(\mathbf{x}),\mathrm{\dots},\nabla {g}_{m}(\mathbf{x})\}$ is a set of linearly independent vectors on all points where the constraints are verified.

Note that this is equivalent^{} to finding the stationary points of:

$$f(\mathbf{x})-\sum _{i=1}^{m}{\lambda}_{i}({g}_{i}(\mathbf{x}))$$ |

for $\mathbf{x}$ in the domain and the *Lagrange multipliers* ${\lambda}_{i}$ without restrictions^{}.

After finding those points, one applies the ${g}_{i}$ constraints to get the actual stationary points subject to the constraints.

Title | Lagrange multiplier method |
---|---|

Canonical name | LagrangeMultiplierMethod |

Date of creation | 2013-03-22 12:25:10 |

Last modified on | 2013-03-22 12:25:10 |

Owner | cvalente (11260) |

Last modified by | cvalente (11260) |

Numerical id | 10 |

Author | cvalente (11260) |

Entry type | Definition |

Classification | msc 49K30 |

Related topic | ExampleOfCalculusOfVariations |

Related topic | IsoperimetricProblem |

Defines | Lagrange multiplier |