Suppose that $f(x,y)$ is a two variable function and we want to find its maximum that is
\begin{align}\max_{(x,y)}f(x,y)\end{align}
where $(x,y)\in(-\infty,+\infty)\times(0,+\infty)$. The right path, to find it, is to take the partial derivaves with respect to $y$ and $x$ and form the first order conditions (FOC) we obtain that $x^*=x(y)$ and $y^*=y(x)$ i.e. $x^*$ and $y^*$ are functions of $y$ and $x$ respectively. Then by solving the system of $(x^*,y^*)$ we have the possible solutions of the problem. To ensure that the function has a maximum we need to check the second order conditions with the Hessian matrixe, i.e. the concavity condition. Although, in my problem I have that the second oreder derivative with respect to $x$ is negative, i can not find the sign of the second order derivative with respect to $y$ and by taking the Hessian matrix I am not sure about the sign of the determinant of the whole $2\times2$ matrix. So another way to solve the problem is to find the $x^*=x(y)$ from the FOC with respect to $x$ and then to substitute this in the function to obtain $$g(y)=f(x^*(y),y)$$ and then I solve for (the equivalent problem)
\begin{align}\max_{y}g(y)\end{align} In this case, I show that, since $y\in(0,+\infty)$, the function $g$ is increasing and concave with respect to $y$ and uniformly bounded from above, which means that in order to find the maximum, I must let $y\to+\infty$ (it has a supremum). Instead of finding an explicit solution of (x,y), I have sowhn that $x^*=x(y)$ and that $g$ which comes from the substitution of $x=x^*$ to the intitial function $f$ gives a maximum only if $y\to+\infty$.
My questions are
- Is my reasoning in the right paths, since I can not change the restrictions that I have for my function, do I need anything more to complete my proof?
- Has anybody seen any such approach of a maximum problem in any textbook or paper? I am trying to define it rigrously from a mathematic point of view.
P.S. I can not provide any mathematical formulas or solutions due to the complexity of the problem.