proof of calculus theorem used in the Lagrange method

Let f(𝐱) and gi(𝐱),i=0,,m be differentiableMathworldPlanetmathPlanetmath scalar functions; 𝐱Rn.

We will find local extremes of the function f(𝐱) where f=0. This can be proved by contradictionMathworldPlanetmathPlanetmath:


but then f(𝐱) is not a local extreme.

Now we put up some conditions, such that we should find the 𝐱SRn that gives a local extreme of f. Let S=i=1mSi, and let Si be defined so that gi(𝐱)=0𝐱Si.

Any vector 𝐱Rn can have one componentPlanetmathPlanetmathPlanetmath perpendicularMathworldPlanetmathPlanetmath to the subset Si (for visualization, think n=3 and let Si be a flat surface). gi will be perpendicular to Si, because:


But gi(𝐱)=0, so any vector 𝐱+ϵgi must be outside Si, and also outside S. (todo: I have proved that there might exist a component perpendicular to each subset Si, but not that there exists only one; this should be done)

By the argumentMathworldPlanetmath above, f must be zero - but now we can ignore all components of f perpendicular to S. (todo: this should be expressed more formally and proved)

So we will have a local extreme within Si if there exists a λi such that


We will have local extreme(s) within S where there exists a set λi,i=1,,m such that

Title proof of calculus theorem used in the Lagrange method
Canonical name ProofOfCalculusTheoremUsedInTheLagrangeMethod
Date of creation 2013-03-22 13:29:51
Last modified on 2013-03-22 13:29:51
Owner mathcam (2727)
Last modified by mathcam (2727)
Numerical id 6
Author mathcam (2727)
Entry type Proof
Classification msc 15A18
Classification msc 15A42