# About Implicit Function Theorem and Lagrange Multipliers

I am studying the meaning of the multiplier in Lagrange Multiplier Method and got the following question. I tried it myself, but I am not sure if it is correct. I would really appreciate it if someone could help me check. In order to state the question, we need some background information first:

Background Information

Theorem$$\space\space\space\space$$ Let $$f$$ and $$h$$ be $$C^1$$ functions of two variables. For any field value of the parameter $$a$$, let $$(x^*(a), y^*(a))$$ be the solution of the following maximization problem $$$$Max\space\space\space\space f(x, y)\\ \space\space\space\space\space\space\space\space\space s.t.\space\space\space\space h(x, y) = a$$$$ with corresponding multiplier $$\mu^*(a)$$. Suppose that $$x^*$$, $$y^*$$, and $$\mu^*$$ are $$C^1$$ functions of $$a$$ and that NDCQ holds at $$(x^*(a), y^*(a), \mu^*(a))$$. Then, $$$$\mu^*(a) = \frac{d}{da}f(x^*(a),y^*(a)).$$$$

Question

Use the Implicit Function Theorem to write out a specific inequality which would guarantee the validity of the assumption in the above theorem that the solution $$(x(a), y(a))$$ of the maximization problem depends smoothly on $$a$$.

My Attempt

Consider the maximization problem as in the above theorem: $$$$Max\space\space\space\space f(x, y)\\ \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space s.t.\space\space\space\space h^*(x, y; a) \equiv h(x, y) - a = 0.$$$$ Assume that the constraint qualification holds at $$(x^*(a), y^*(a))$$, the solution to the maximization problem: $$$$\nabla h^*(x, y; a) = \begin{pmatrix} \frac{\partial h^*}{\partial x}(x^*(a), y^*(a); a)\\ \frac{\partial h^*}{\partial y}(x^*(a), y^*(a); a). \end{pmatrix}$$$$ Form the Lagrangian: $$$$L(x, y, \mu; a) = f(x, y, ;a) - \mu h^*(x, y; a).$$$$ The constrained maximizer $$(x^*(a), y^*(a))$$ must satisfy the F.O.C.s: $$$$\frac{\partial L}{\partial x}(x, y, \mu; a) = 0\\ \frac{\partial L}{\partial y}(x, y, \mu; a) = 0\\ \frac{\partial L}{\partial \mu}(x, y, \mu; a) = 0,$$$$ a system of 3 equations in 3 unknowns.
By the Implicit Function Theorem, we need the matrix $$$$\begin{pmatrix} \frac{\partial^2 L}{\partial x^2} & \frac{\partial^2 L}{\partial x \partial y} & \frac{\partial^2 L}{\partial x \partial \mu}\\ \frac{\partial^2 L}{\partial x \partial y} & \frac{\partial^2 L}{\partial y^2} & \frac{\partial^2 L}{\partial y \partial \mu}\\ \frac{\partial^2 L}{\partial x \partial \mu} & \frac{\partial^2 L}{\partial y \partial \mu} & \frac{\partial^2 L}{\partial \mu^2} \end{pmatrix} = \begin{pmatrix} \frac{\partial^2 L}{\partial x^2} & \frac{\partial^2 L}{\partial x \partial y} & \frac{\partial^2 L}{\partial x \partial \mu}\\ \frac{\partial^2 L}{\partial x \partial y} & \frac{\partial^2 L}{\partial y^2} & \frac{\partial^2 L}{\partial y \partial \mu}\\ \frac{\partial^2 L}{\partial x \partial \mu} & \frac{\partial^2 L}{\partial y \partial \mu} & 0 \end{pmatrix},$$$$ which is the Hessian matrix $$D^2L$$ of the Lagrangian, to be nonsingular. Therefore, the inequality we are looking for is that $$$$det(D^2L) \neq 0.$$$$

• What is NDCQ? ${}{}$ Aug 19, 2023 at 22:58
• No, NDCQ is equivalent to LICQ, and It does not mean that the Hessian of the Lagrangian is nonsingular. The Implicit function theorem allows us to see that $(x(a),y(a))$ is differentiable. So, it is only necessary to use the Chain Rule to differentiate the objective function $f(x(a),y(a))$ Aug 26, 2023 at 10:17