Lagrange Multiplier - Handling Multiple Constraints

Handling Multiple Constraints

The method of Lagrange multipliers can also accommodate multiple constraints. To see how this is done, we need to reexamine the problem in a slightly different manner because the concept of “crossing” discussed above becomes rapidly unclear when we consider the types of constraints that are created when we have more than one constraint acting together.

As an example, consider a paraboloid with a constraint that is a single point (as might be created if we had 2 line constraints that intersect). Even though the level set (i.e., contour line) seems to be “crossing” that point and its gradient doesn't seem to be parallel to the gradients of either of the two line constraints, in fact, the relation between that point and the contour lines are similar to relation of a line tangent to a sphere in 3D space. So the point still can be considered "tangent" to the level contours. And the point is obviously a maximum and a minimum because there is only one point on the paraboloid that meets the constraint.

While this example seems a bit odd, it is easy to understand and is representative of the sort of “effective” constraint that appears quite often when we deal with multiple constraints intersecting. Thus, we take a slightly different approach below to explain and derive the Lagrange Multipliers method with any number of constraints.

Throughout this section, the independent variables will be denoted by and, as a group, we will denote them as . Also, the function being analyzed will be denoted by and the constraints will be represented by the equations .

The basic idea remains essentially the same: if we consider only the points that satisfy the constraints (i.e. are in the constraints), then a point is a stationary point (i.e. a point in a “flat” region) of f if and only if the constraints at that point do not allow movement in a direction where f changes value.

Once we have located the stationary points, we need to do further tests to see if we have found a minimum, a maximum or just a stationary point that is neither.

We start by considering the level set of f at . The set of vectors containing the directions in which we can move and still remain in the same level set are the directions where the value of f does not change (i.e. the change equals zero). Thus, for every vector v in, the following relation must hold:

where the notation above means the -component of the vector v. The equation above can be rewritten in a more compact geometric form that helps our intuition:

This makes it clear that if we are at p, then all directions from this point that do not change the value of f must be perpendicular to (the gradient of f at p).

Now let us consider the effect of the constraints. Each constraint limits the directions that we can move from a particular point and still satisfy the constraint. We can use the same procedure, to look for the set of vectors containing the directions in which we can move and still satisfy the constraint. As above, for every vector v in, the following relation must hold:

From this, we see that at point p, all directions from this point that will still satisfy this constraint must be perpendicular to .

Now we are ready to refine our idea further and complete the method: a point on f is a constrained stationary point if and only if the direction that changes f violates at least one of the constraints. (We can see that this is true because if a direction that changes f did not violate any constraints, then there would a “legal” point nearby with a higher or lower value for f and the current point would then not be a stationary point.)