Extrema with Constraints II
As we said last time, we have an idea for a necessary condition for finding local extrema subject to constraints of a certain form. To be explicit, we assume that is continuously differentiable on an open region , and we also assume that is a continuously differentiable vector-valued function on the same region (and that ). We use to define the region consisting of those points so that . Now if is a point with a neighborhood so that for all we have (or for all such ). And, finally, we assume that the determinant
is nonzero at . Then we have reason to believe that there will exist real numbers , one for each component of the constraint function, so that the equations
are satisfied at .
Well, first off we can solve the first equations and determine the right off. Rewrite them as
a system of equations in the unknowns . Since the matrix has a nonzero determinant (by assumption) we can solve this system uniquely to determine the . What’s left is to verify that this choice of the also satisfies the remaining equations.
To take care of this, we’ll write , so we can write the point as and particularly . Now we can invoke the implicit function theorem! We find an -dimensional neighborhood of and a unique continuously differentiable function so that and for all . Without loss of generality, we can choose so that , where is the neighborhood from the assumptions above.
This is the parameterization we discussed last time, and we can now substitute these functions into the function . That is, we can define
Or if we define we can say this more succinctly as and .
Anyhow, now all of these are identically zero on as a consequence of the implicit function theorem, and so each partial derivative is identically zero as well. But since the are composite functions we can also use the chain rule to evaluate these partial derivatives. We find
Similarly, since has a local minimum (as a function of the ) at we must find its partial derivatives zero at that point. That is
Now let’s take the previous equation involving , evaluate it at , multiply it by , sum over , and add it to this latest equation. We find
Now, the expression in brackets is zero because that’s actually how we defined the way back at the start of the proof! And then what remains is exactly the equations we need to complete the proof.