Does it matter which function you use as a constraint, and which function you use as your main function when optimizing a function using the method of Lagrange multipliers? For example: Let C be the curve of intersection between $z = 4 - x^2 - y^2$ and $4x^2y = 1$? Find the biggest value of $z$ on this curve for $x>0$. Can I use either as my main function, and either as my constraint, or is it important which is which? If so, why?
Thanks in advance
Your problem can be stated,
$$\max\limits_{x,y} 4-x^2-y^2\\ \mbox{such that:}\quad4x^2y=1, \quad x>0.$$
The objective function is $4-x^2-y^2$ and the constraints are $4x^2y=1$ and $x>0$.
It matters which is the objective and which are the constraints, they are fundamentally different unless you write the constraints are indicator functions. It doesn't make sense to try to maximize or minimize $4x^2y=1$, the left hand side is always $1$ by definition.
We can write the lagrangian for our problem in the following way:
$$L(x,y,\lambda,\mu) = 4-x^2-y^2 +\lambda (4x^2y-1) +\mu x$$
with Lgrange multipliers $\lambda\in\mathbb{R}$ and $\mu<0$.