Why can we simply substitute the constrain in when maximizing this equation? I am working on some Portfolio Analysis related material at the moment, and am trying to understand why the following approach to solving this maximization problem is correct:
$\max _x \frac{\mu' x-r_f}{(x' \varSigma x)^{0.5}} $ subject to $\mathbf{1}'_Nx=1$,
where $\varSigma$ is a positive definite $N\times N$ matrix, $r_f \in \mathbb{R}$, and $x,\mu \in \mathbb{R}^N$
To solve this, we substitute the add-up constrain into the objective function
$\frac{\mu' x-r_f}{(x' \varSigma x)^{0.5}} $ to obtain an unconstraint maximization problem. Because $\mathbf{1}_N'x=1$, we can write $r_f = r_f \mathbf{1}_N'x$. Combining this the objective function becomes $\beta = \frac{\mu' x-r_f\mathbf{1}_N'x}{(x' \varSigma x)^{0.5}} $. Now what they do is find the zero-point of the derivative of this function, and conclude that this must be the maximum. However it is not clear to me why a possible zeropoint of the $\beta$ function would fullfill the condition $\mathbf{1}'_Nx=1$ and why it should necessarily be a maximum. Is there something I am misssing here or is this just a very shoddy proof? If so, how would one go about proving correctly?

There is no need for it to sum up to $1$. The problem is homogeneous in $x$ meaning that if you replace $x$ with $tx$, the objective is the same. Hence, once you have a solution to the problem, you can always scale it suitably afterwards. The only important thing is the relative size of the elements in the allocation.
