Condition for increase in the optimum of a general function For a function $f(x,y)$ with the following properties:


*

*$f(x,y)$ is strictly increasing as a function of $x$

*$f(x,y)$ is strictly decreasing as a function of $y$

*$\lim_{x\to\infty}\frac{\partial f(x,y)}{\partial x}=0$

*$\lim_{y\to\infty}\frac{\partial f(x,y)}{\partial y}=0$


I'm studying the optimum of $\frac{f(x,y)}{x}$ with respect to $x$, denoted $x_0$:
$$
Q=\frac{\partial (f(x,y)/x)}{\partial x}\bigg|_{x=x_0}=0.
$$
Assuming $x_0$ is an optimum, I wish to study how $x_0$ changes as $y$ changes, and I arrive at the following
$$
\frac{dx_0}{dy}=-\frac{\partial Q/ \partial y}{\partial Q/ \partial x_0}.
$$
Given that $x_0$ is an optimum, the denominator of the above will be negative and hence the sign of $\frac{dx_0}{dy}$ will be the same as the sign of $\frac{\partial Q}{\partial y}$. And we have the following
$$
sgn\bigg(\frac{\partial Q}{\partial y}\bigg)=sgn\bigg(x_0\frac{\partial (\partial f(x_0,y)/\partial x_0)}{\partial y}-\frac{\partial f(x_0,y)}{\partial y}\bigg).
$$ 
Can we say that the following will always be true?
$$
\frac{\partial Q}{\partial y}\neq 0
$$
In other words, if $y$ changes, the optimum $x_0$ must also change.
If not, can we say something about the cases where this will not be true? Thanks a lot for your help.
 A: I assume the optimum $x^*$ you are looking for corresponds to a maximization problem:
$$x^* = \arg\max_x \frac{ f(x,y) }{ x } = \arg\max_x[ \log f(x,y)  - \log x]$$
Notice that the second equality is true since taking $\log$ preserves the $\arg \max$.
The analysis you are trying to do by defining the function $Q$ was done by The Topkis theorem (http://en.wikipedia.org/wiki/Topkis%27s_theorem) provides conditions for the monotonicity of the $\arg \max$ with respect to the parameter $y$. In order for the optimum $x^*$ to increase with $y$, we need that the quantity inside the $\arg \max$ is supermodular with respect to $x$ and $y$ (http://en.wikipedia.org/wiki/Supermodular_function).
This means that for all $x_1 \le x_2$ and $y_1 \le y_2$ we must have:
$$\log f(x_1,y_1) + \log f(x_2,y_2) \ge \log f(x_2,y_1) + \log f(x_1,y_2)$$
which is the same as:
$$\frac { f(x_2,y_1) } {f(x_2,y_2)} \le \frac { f(x_1,y_1) } {f(x_1,y_2)}$$
This means that the ratios $\frac { f(x,y_1) } {f(x,y_2)}$ are decreasing with $x$. An alternative formulation when $f$ is differentiable is:
$$\frac {\partial \log f(x,y) } {\partial x \partial y} \ge 0 \rightarrow f_{1,2}(x,y) f(x,y) \ge f_1(x,y) f_2(x,y)$$
The conditions you gave don't seem to imply anything about the truth of the inequality above but I couldn't construct a counterexample because I couldn't figure out what the two limits imply. Also, it is a bit unclear what domain you want $f$ to be defined in. Since you are dividing by $x$, is it that $x$ needs to be greater than $0$, or do you also need $f(0,y) = 0$?
I suggest you look into Topkis theorem and see if you can apply it to your problem.
