For shake of simplicity I will consider an easy example: $$\min{x^2-x} {\ {\ }} s.t.{\ {\ }} x-3\geq 0$$ The solution of this problem is quite simple. x=3, and using lagrange multipliers we would obtain a lagrange multiplier value of $\lambda=5$
However, supose im interested on finding the dual. As I have read so far, firstly I would need to find the Lagrangian, in my case $$L(x,\lambda) = x^2-x-\lambda(x-3)$$ Then, I obtain the first derivative $$\frac{\partial L(x,\lambda)}{\partial x} = 2x-1-\lambda$$ From here, I obtain an expresssion of x in terms of the lagrange multiplier, $\lambda$: ${\ } x=\dfrac{\lambda+1}{2}$
So now I can substitute this value in the lagrangian obtaining an expression just in terms of the lagrange multiplier: $$L(\lambda) = (\dfrac{\lambda+1}{2})^2-\dfrac{\lambda+1}{2}-\lambda(\dfrac{\lambda+1}{2}-3)$$
And this equation would be, theoretically, the dual problem that I should maximize subject to the restriction of $\lambda \geq 0$. However, this expression leads to two possible $\lambda$ values: $\lambda = 5 -2\sqrt{6}$ or $\lambda=5+2\sqrt{6}$, both of them far away from the expected $\lambda=5$ value.
This makes me think that I am doing something wrong, but I have no idea of what could it be. I would appreciate any hint you can give me.
$\lambda = 5 \pm 2 \sqrt{6}$ would be from solving $L(\lambda) = 0$, not $L'(\lambda) = 0$.