Is there any technique to deal with a problem where we have a linear objective function and one or many quadratic non-convex function(s) like the problem below?
$$\begin{array}{ll} \text{minimize} & c_1 x + c_2 y\\ \text{subject to} & xy = c\end{array}$$
where $c$, $c_1$ and $c_2$ are arbitrary real constants.
You can apply the Lagrange multiplier method.
$\mathcal L=c_1x+c_2y+\lambda[c-xy]$
$\frac{\partial \mathcal L}{\partial x}=c_1-\lambda y=0\Rightarrow c_1=\lambda y\quad (1)$
$\frac{\partial \mathcal L}{\partial y}=c_2-\lambda x=0\Rightarrow c_2=\lambda x \quad (2)$
$\frac{\partial \mathcal L}{\partial \lambda}=c-xy=0\Rightarrow c=xy \quad (3)$
Dividing (1) by (2):
$\frac{c_1}{c_2}=\frac{y}{x}\Rightarrow y=\frac{c_1}{c_2}x$
Inserting the term for $c$ in $(3)$
$c= \frac{c_1}{c_2}x^2$
$x_1= \sqrt{c\frac{c_2}{c_1}}, x_2=-\sqrt{c\frac{c_2}{c_1}}$
$\Rightarrow y_1=\frac{c_1}{c_2} \sqrt{c\frac{c_2}{c_1}}, y_2=-\frac{c_1}{c_2} \sqrt{c\frac{c_2}{c_1}}$
The two stationary points $(x_1/y_1)$ and $(x_2/y_2)$ can be local maximums or local minimums. Insert the corresponding values into the $\texttt{bordered Hessian}$ to evaluate what kind of stationary points the are.
$$\tilde H=\left( \begin{array}{} 0 & \frac{\partial^2 \mathcal L}{\partial \lambda\partial x}& \frac{\partial^2 \mathcal L}{\partial \lambda\partial y} \\ \frac{\partial^2 \mathcal L}{\partial \lambda\partial x} & \frac{\partial^2 \mathcal L}{\partial x\partial x} & \frac{\partial^2 \mathcal L}{\partial x\partial y} \\ \frac{\partial^2 \mathcal L}{\partial \lambda\partial y} & \frac{\partial^2 \mathcal L}{\partial x\partial y} & \frac{\partial^2 \mathcal L}{\partial y\partial y} \end{array}\right)$$
If $det \ \tilde H(x_0,y_0) >0 \Rightarrow \texttt{it´s a (local) maximum}$
If $det \ \tilde H(x_0,y_0) <0 \Rightarrow \texttt{it´s a (local) minimum}$
