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...

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

1 Answer 1


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.

  • $\begingroup$ 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? $\endgroup$ May 5, 2020 at 19:05
  • $\begingroup$ Yes, first you find $a$ then $(x,y)$. $\endgroup$
    – Reda
    May 6, 2020 at 3:47
  • $\begingroup$ 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 $y$. $\endgroup$ May 6, 2020 at 15:43
  • $\begingroup$ 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. $\endgroup$
    – Reda
    May 6, 2020 at 17:27
  • $\begingroup$ Could you please verify these textbooks, about optimization? $\endgroup$ May 6, 2020 at 19:03

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