Let $A, B, C, D \in \mathbb{R^\*\_+}$.


Is it possible to solve
$$
\max\_{ \substack{0 \leq x\leq A \\ 0\leq y\leq B}} \frac{1+x+y}{(1+Cx)(1+Dy)}
$$


The KKT conditions give for an extrema $(x^\*,y^\*)$


If $\frac{1}{C} > (1+y^\*)$ then $x^\*=A$ otherwise $x^\*=0$,


If $\frac{1}{D} > (1+x^\*)$ then $y^\*=B$ otherwise $y^\*=0$.


Thus it would solve the problem (if one could show concavity in $(x,y)$) when $1/C>1+B$ and $1/D>1+A$.


I have no idea for the case $1/C\leq 1+B$ or $1/D\leq 1+A$



Optimization function of two variables
The gradient is $0$ iff $[x,y] = [1/D-1, 1/C-1]$, at which point the objective value is $1/(C+D-CD)$. If this is in the feasible region $[0, A] \times [0,B]$, it may be optimal. Compare to the optimal values on the edges $\{0\} \times [0,B]$, $\{A\} \times [0,B]$, $[0,A] \times \{0\}$, $[0,A] \times \{B\}$.


