Lagrange method: case of many equality constraints
Consider the problem
(1) maximize subject to
(the number of constraints is smaller than the number of variables ).
The idea is to understand the proof in case of one constraint and to develop the format of the result for many constraints by analogy.
Implicit function theorem
We saw previously that the condition
is sufficient for existence of the implicit function defined by This statement is asymmetric. What if we want to find as a function of ? The condition
is more general than (2). If , then the restriction defines as a function of and if , then can be found as a function of For example, for a circle for we can find and for we can find and the whole circle is covered.
To obtain the multi-dimensional generalization of (3), denote
The last equation follows from our notation of the gradient (1) becomes
maximize subject to
The required generalization of (3) can now be stated as follows:
(4) The matrix has rank or, equivalently, the vectors are linearly independent.
Remark. The best way to check this condition is to see that .
First and second order conditions
Denote (this is a row vector with as many components as restrictions). The Lagrangian is defined by
The first-order conditions become
The Hessian is
Recall that the refined sufficient condition was based on the set Now it becomes
Finally, we can state the sufficient condition:
Theorem. Assume the implicit function existence condition (4) and consider a critical point for the Lagrangian (that satisfies (5)). a) If at that point for any nonzero , then that point is a minimum point. b) If at that point for any nonzero , then that point is a maximum point.