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I have a function given below:

$f(x,y) = \log_2(1 - xe^y + c + ax + axe^y)$

I was hoping to find $(x,y)$ that minimizes this function. The constraints are $0 < x <1$ and $y > 0$. Here, both $x$ and $y$ are discrete. I wasn't sure if the function is convex or not. So, I tried checking and I think the function is $quasiconcave$. How can I find $(x,y)$ in this case such that it minimizes $f$?

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  • $\begingroup$ whether the problem has a solution will depend on the value of $a$. Is $a<1$? $\endgroup$
    – pegasus
    Mar 4, 2020 at 16:48
  • $\begingroup$ yes, $a < 1$. Does it have a solution? $\endgroup$
    – Bikas
    Mar 4, 2020 at 16:50
  • $\begingroup$ Yes, it does if $x$ and $y$ are continuous. I am not sure, though, what values you are allowed to give to $x$ and $y$ since you mention that they take discrete values. $\endgroup$
    – pegasus
    Mar 4, 2020 at 16:52
  • $\begingroup$ x can be between $0$ and $1$ and y can be greater than $0$ $\endgroup$
    – Bikas
    Mar 4, 2020 at 16:53
  • $\begingroup$ Do you mean $x\in[0,1]$ and $y\geq 0$ (you have strict inequalities in the question). But, then, the problem is not discrete. $\endgroup$
    – pegasus
    Mar 4, 2020 at 16:54

1 Answer 1

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Hint: Notice that $\log_2$ is a strictly increasing function. The same holds for $x\mapsto x+c$.

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  • $\begingroup$ Are you implying that the $(x,y)$ that minimizes $f$ is a corner solution? As $\log_2$ is a strictly increasing function, the solution would be the starting point? $\endgroup$
    – Bikas
    Mar 4, 2020 at 16:46
  • $\begingroup$ Well... it should be the solution that minimizes the argument. How does the expression "inside" the $\log_2$ behave? $\endgroup$
    – pegasus
    Mar 4, 2020 at 16:48
  • $\begingroup$ I think the expression inside the $\log_2$ is not strictly increasing. It increases and then decrease $\endgroup$
    – Bikas
    Mar 4, 2020 at 16:52
  • $\begingroup$ As a function of $x$ or of $y$? Try to factor out what you can. $\endgroup$
    – pegasus
    Mar 4, 2020 at 16:53

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