Optimization problem is to find $x$>0 which
$min \ \ L=\frac{A\left ( B(\frac{C}{Cx-B}+\frac{1}{x})+2C\log(\frac{B}{x}-C) \right )}{B^3}$
$s.t \ \ x\leq K $
Rewriting the objective function using Lagrange multiplier $\lambda$
$J = L + \lambda (x-K)$
While solving the optimization problem with Lagrange, I came to the following equations:
$\frac{\partial J}{\partial x}= \frac{-A}{(Bx-Cx^2)^2}+\lambda$
$\frac{\partial J}{\partial \lambda} = x-K$
Consider that $A,B,C,K>0$ and also $B>Cx$.
After solving $\frac{\partial J}{\partial x}=0$, I am getting $x=\frac{B\pm\sqrt{B^2-4C\sqrt{A}}}{2C}$. But it may not be less than $K$. How to solve this? Please help.