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
(2)
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
(3)
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.
In case of a more general implicit function existence condition (3) our derivation of the necessary and sufficient conditions goes through (if holds, instead of
, one can simply swap
and
).
To obtain the multi-dimensional generalization of (3), denote
and
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
(5)
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.
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