Maximization problem of a quadratic and concave function

Suppose that $$f(\cdot)$$ is a quadratic and concave function such that it has a maximum, $$x \in \mathbb{R}$$, $$y\in(0,+\infty)$$. I have to solve a maximization problem that is $$\max_{\dfrac{x}{y}}f(x/y)$$ My question is that, since I want to maximize with respect to $$x/y$$, I will treat this as if $$\frac{\partial f(x/y)}{\partial (x/y)}$$?

Or do i need to maximize with respect to $$x$$, then in $$y$$ and use the Hessian matrix?

Or is it something else? Well it is a composition of functions after all, isn't it? I am a llitle confused...

• just pick any value for $y$ and then solve for $x$ May 2, 2020 at 12:19
• What you mean when you say pick any value for $y$". Should I treat $y$ as a constant? May 2, 2020 at 23:34
• well, if $(x,y)$ is a solution, then $(2x,2y)$ is also a solution, right? May 3, 2020 at 1:36
• Well I guess so...but I can not understand what is the intuition behind this. How is this going to help me? May 3, 2020 at 7:25

Suppose $$a$$ is the point where $$f$$ is maximal, then, all the couples $$(x,y)$$ that verify $$x/y=a$$ are solutions to the optimization problem.
• So, you mean that, by setting the factor $x/y=a$, I can re-state the maximization problem with respect to this $a$ as the control variable? May 5, 2020 at 19:05
• Yes, first you find $a$ then $(x,y)$.
• Well, does it matter if $a$ is indepedent of $x$ and $y$? What deos this mean about the optimal $x/y$? Or this is the case, if my problem is well defined, then I need to find such an $a$, that is indepedent of the $x$ and $y$. May 6, 2020 at 15:43
• There's some good textbooks about optimization that you can read. For your problem, if $f(x) = \alpha /2 x^2+\beta x +\gamma$, the optimum is at $a=-\beta/\alpha$. Therefore, all the couples $(x,y)$ that verify $x=ay$, for $y$ nonnegative, are solutions to the optimization problem.