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I am trying to solve this optimization problem, x, y are possibilities, and $x,y \in [0,1]$, $a, b , D_1, D_2, $are known given parameters.$D_1 +D_2 = D$

$$minimize \qquad f(x,y) = axy$$ \begin{align} s.t. -xlnx + bx - D_1 &\leq 0 \\ -ylny + by - D_2 &\leq 0 \\ x-1 &\leq0\\ y-1 &\leq0 \end{align}

because of the $ln(x)$ definition, $x,y$ can not be 0. I only add the $x,y \leq 1$ to the constriant.

And I have tried the KKT conditions to solve the problem. \begin{align} &\mathcal{L}(x,y,\lambda_1,\lambda_2,\lambda_3) = axy +\lambda_1(-xlnx+ bx -ylny + by - D)+\lambda_2(x-1)+\lambda_3(y-1)\\ &\frac{\partial\mathcal{L}}{\partial x} = ay+\lambda_1(-lnx-1+b) +\lambda_2=0\\ &\frac{\partial\mathcal{L}}{\partial y} = ax+\lambda_1(-lny-1+b) +\lambda_3= 0\\ &-xlnx+ bx -ylny + by - D = 0\\ & x-1 =0\\ & y -1 = 0 \end{align}

Actually I have an assumption that the function get the minimum when $D_1 = D_2$, also means $x = y$

Suppose only the first constraint is satisfied, $-xlnx+ bx -ylny + by - D = 0$, I can only got one equation $$-xlnx +bx -x = -ylxy +by -y$$

which equals to$$D_1-x=D_2-y$$

I am not sure should I use KKT condition here because it looks like cannot found the primal solution, could you help me solve the optimization problem? Thanks ahead!

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