# convex optimization with KKT conditions minimize f(x,y) = axy

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!