The Lagrangian Functional   

If we define a function of the three variables x, y, and l  by

L( x,y,l) = f( x,y) -l( g( x,y) -k)

then the critical points of L( x,y,l) occur when

¶L ¶x = 0,         ¶L ¶y = 0,         ¶L ¶l = 0

However, L_x = f_x-lg_x, L_y = f_y-lg_y, and L_l =  g( x,y) -k. Thus, the critical points of L( x,y,l) are solutions of the system of equations

fx = lgx,        fy = lgy,    g( x,y) = k

(3)

That is, the Lagrange multiplier method (1) is equivalent to finding the critical points of the function L( x,y,l).  

The function L( x,y,l) is called a Lagrangian of the constrained optimization.  Lagrangians allow us to extend the Lagrange multiplier method to functions of more than two variables. Let's revisit a problem from the previous section to see this idea at work.

EXAMPLE 6  Find the point(s) on the plane x+y-z = 3 that are closest to the origin.

Solution: To begin with, we let f denote the square of the distance from a point ( x,y,z) to the origin. Consequently,

f = x2+y2+z2

Thus, we want to 

minimize f(x,y,z)  = x² + y² +z²

subject to  x + y - z = 3

To do so, we define the Lagrangian of the problem to be

 L( x,y,l) = x² + y² +z² -l( x+y-z - 3)

Since L_x = 2x-l  ,L_y = 2y-l, L_z= 2z+l, and Ll = -(x+y-z - 3), the critical points of L are the solutions to

2x=l  ,  2y=l,  -2z=l,   and  x + y - z = 3

Since 2x+ 2y - 2z = 6, the other three equations imply that

l+ l +l = 6,   or l = 2

However, l = 2 implies that x=1, y=1, and z=-1, which means that (1,1,-1) is the point on x+y-z = 3 closest to the origin. 

We can also use (3) to find the extrema of a function f( x,y,z) subject to two constraints,

g( x,y,z) = k,        h( x,y,z) = l

In particular, we define a Lagrangian function of the 3 variables x,y,and z, and the 2 auxiliary variables l and m by

L( x,y,z,l,m) = f( x,y,z) -lg1( x,y,z) -mh1( x,y,z)

where g₁( x,y,z) = g( x,y,z) -k and h₁(x,y,z) = h( x,y,z) -l. It then follows that the partial derivatives of L are

¶L ¶x = fx( x,y,z) -lgx( x,y,z) -mhx( x,y,z) ,  ¶L ¶y = fy( x,y,z) -lgy( x,y,z) -mhy( x,y,z) ,  ¶L ¶y = fy( x,y,z) -lgz( x,y,z) -mhz( x,y,z) ,

¶L ¶l = g( x,y,z) ,     ¶L ¶m = h( x,y,z)

Thus, the critical points of L lead to the Lagrange multiplier problem

fx = lgx+mhx,      fy = lgy+mhy,   fz = lgz+mhz

subject to the constraints

g( x,y,z) = 0        and        h(x,y,z) = 0

In the same fashion, this method generalizes to any number of variables subject to a large number of constraints, as is discussed in the accompanying worksheet.