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# Lagrange's multipliers.

Proposition 6.25   Let f(x,y,z) be a function and let S be a surface whose equation is given by g(x,y,z)=0. The extrema of f on the surface S are afforded at the points for which there exists a scalar such that

The same proposition is valid for a function of two variables, when the constraint is given by a curve in the plane.

Example 6.26

Example 6.27   Let f(x,y,z)=x2+2y2+z2 and let S be the cylinder given by x2+4z2=1, i.e. we consider the function g(x,y,z)= x2+4z2-1. Then, we have:

We have:

For which values of do these equations have a solution?

If x=0, then and we found two points and .

If , then z=0 and and this satisfies the first equation too. We get two points P3(1,0,0) and P4(-1,0,0).

Which of these points correpsond to minima and which of them correspond to maxima? We compute the values:

• ; at these points f has a minimum.
• f(P3)=f(P4)=1; at these points f has a maximum.

Proposition 6.28   Let f(x,y,z) be a function and let C be the intersection of two surfaces whose equation are given by g1(x,y,z)=0 and g2(x,y,z)=0. The extrema of f on the curve C are afforded at the points for which there exist scalars and such that

Example 6.29   The plane cuts the cylinder x2-y2=1 (displayed on Fig.  7(a)) along a curve (one branch is displayed on Fig.  7(b)).

Find the points on at the shortest distance from the origin. This means that we consider the function f(x,y,z)=x2+y2+z2 with the constraints g1(x,y,z)=x-y+x-1 and g2(x,y,z)=x2-y2-1.