When I write the K.K.T. conditions for the problem I have, I get the following expression for the gradient of the Lagrangian:
$$\frac{\partial \mathcal{L}}{\partial x} = - \frac{\sqrt{x} + \sqrt{y}}{\sqrt{x}\left(1+(\sqrt{x}+\sqrt{y})^2\right)} + A = 0$$
$$\frac{\partial \mathcal{L}}{\partial y} = - \frac{\sqrt{x} + \sqrt{y}}{\sqrt{y}\left(1+(\sqrt{x}+\sqrt{y})^2\right)} + B = 0$$
Where $A$ and $B$ are the functions of the corresponding Lagrangian multipliers only.
Since Lagrangian partial derivatives are not defined for $x=0$ and $y=0$, can I say that therefore $x \neq 0$ and $y \neq 0$ for the optimal solution. In other words, I need to have strict inequalities $x > 0$ and $y > 0$ instead of $x \geq 0$ and $y \geq 0$ at the optimum point.
Thank you very much in advance, Lazar
I provide the complete set of K.K.T. conditions in the sequel:
$\frac{\partial \mathcal{L}}{\partial x_k}, \frac{\partial \mathcal{L}}{\partial y_k} = 0$
$\sum_{k=1}^{n} x_k \leq C_n$ $\text{for}\; n=1\ldots N$
$\sum_{k=1}^{n} y_k \leq D_n$ $\text{for}\; n=1\ldots N$
$x_k,y_k \geq 0$
$\lambda_n, \nu_n, \mu_k, \xi_k \geq 0$
$\lambda_n \left(\sum_{k=1}^{n} x_k - C_n\right) = 0$ $\text{for}\; n=1\ldots N$
$\nu_n \left(\sum_{k=1}^{n} y_k - D_n\right) = 0$ $\text{for}\; n=1\ldots N$
$-\mu_k x_k = 0$ $\text{for}\; k=1\ldots N$
$-\xi_k y_k = 0$ $\text{for}\; k=1\ldots N$
I want to see whether the KKT conditions (derivatives given at the beginning) imply $x \neq 0$ and $y \neq 0$, resulting in $\mu = 0$ and $\xi = 0$ from two equations at the end.
Thank you.