I want to minimize $a_1 x_1+a_2 x_2+c x_1 \log\dfrac{x_1}{x_1+x_2}+c x_2 \log\dfrac{x_2}{x_1+x_2}$ for $x_{1,2}\ge 0$ (all the scalars $a_{1,2}<0$ and $c>0$ are real).
After having trouble solving for the critical points of the Lagrangian, I thought I should check its second order derivatives, only to figure out that the Hessian matrix has a determinant that vanishes to zero.
How am I supposed to perform the optimization when I can't find critical points and even if I could, I cannot verify their status (max, min or saddle point)?
First Derivatives
The first order derivatives of the function are: $a_j+c\log\dfrac{x_j}{x_1+x_2}$, for $j=1,2$.
I've noticed that, if either of the $x$'s becomes $0$ (achieves its lower feasible bound) then the derivative for that $x$, goes to minus infinity. This seems to imply to me that the origin is not a candidate.
Second Derivatives
The Hessian of the objective function is given by $\left(
\begin{array}{cc}
\text{c} \left(\frac{1}{x_1}-\frac{1}{x_1+x_2}\right) &
-\frac{\text{c}}{x_1+x_2} \\
-\frac{\text{c}}{x_1+x_2} & \frac{\text{c} x_1}{x_2(x_1+x_2)}
\\
\end{array}
\right)$ and has a determinant equal to zero.
The question
How should I conceptualize this problem? Is there something I'm missing? Where can I find info on how to tackle functions like this in optimization problems?
Best Answer
It only makes sense to consider critical points in the interior $\Omega$ of the first quadrant. Let $c=1$. The equations for the critical points are $${x_1\over x_1+x_2}=e^{-a_1},\qquad {x_2\over x_1+x_2}=e^{-a_2}\ .$$ They only have solutions if $$e^{-a_1}+e^{-a_2}=1\ .\tag{1}$$ If this is not the case there is no critical point in $\Omega$. On the other hand, if $(1)$ is satisfied then there is a full ray ${x_2\over x_1}={\rm const.}$ consisting of critical points. These critical points then are degenerate (the given function is even constant along this ray), and the Hessian is of no help.
In fact your function is homogeneous of degree $1$. Write $$x_1+x_2=:s\geq0,\qquad x_1=(1-t)s,\quad x_2=t s\quad(0\leq t\leq1)\ ,$$ and obtain a function $$g(s,t):=f\bigl((1-t)s,,ts\bigr)=s\> G(t)\ .$$