Stuck on step in Lagrangian Problem For $w,E$ column vectors, $i$ the vector of ones, and $\Sigma$ - an $n\times n$ positive definite symmetric matrix, I am trying to solve the following maximization problem:
$$
\max_{\{ w\}} \left\{ \frac{w^T E}{\sqrt{w^T\Sigma w}}\right\} \quad s.t. \quad w^T i = 1
$$
I form the lagrangian:
$$
L = \frac{w^T E}{\sqrt{w^T\Sigma w}}+\lambda(w^T i -1)
$$
$$
\frac{dL}{dw}= \frac{E(w^T\Sigma w)^{1/2} - w^TE \Sigma w (w^T\Sigma w)^{-1/2} }{w^T\Sigma w} + \lambda i \overset{\text{set}}{=} 0
$$
but I'm confused how to proceed, usually we put the derivative in terms of $w$ and use the constraint, but that's not so easy in this case, any hints?
update Instead solving the following problem as proposed in the comments:
$$
\max_{\{w\}} w^T E \quad s.t. \quad w^Ti=1, w^T\Sigma w=\sigma^2
$$
leads to the result:
$$
w = \Sigma^{-1} \frac{E (C\mu -B)+1(A-B\mu)}{AC-B^2}
$$
$$
\implies w^T \Sigma w = \frac{C\mu^2-2B\mu + A}{AC-B^2}
$$
where:
$$
A = E^T \Sigma^{-1}E \quad~~~~~~ B = E^T \Sigma^{-1}1 ~~~~~\quad  C= 1^T \Sigma^{-1}1
$$
 A: Let $\Sigma^{1/2}$ be a square root of $\Sigma$ and perform the variable change
$$
u=\Sigma^{1/2}w,\quad a=\Sigma^{-1/2}E,\quad b=\Sigma^{-1/2}i.
$$
The problem becomes
$$
\max\frac{a^Tu}{\|u\|}\quad\text{subject to } \ b^Tu=1.
$$
Since the objective function depends only on the unit vector $\frac{u}{\|u\|}$ we can replace the condition $b^Tu=1$ with $b^Tu>0$ and normalize $u$ to $b^Tu=1$ at the end. We have the following cases
Case 1: $a^Tb>0$ (i.e. $a$ and $b$ pointing at the same half-space). Then the maximum is clearly attained at the same direction as $a$, i.e. $u_0\|a$ and
$$
\max=\frac{a^Ta}{\|a\|}=\|a\|.
$$
Normalization gives $u_0=\frac{a}{b^Ta}$.
In the original notations: $w_0=\Sigma^{-1/2}u_0=\frac{\Sigma^{-1}E}{i^T\Sigma^{-1}E}$ and $\max=\sqrt{E^T\Sigma^{-1}E}$.
Case 2: $a^Tb\le 0$ (i.e. $a$ and $b$ pointing at the opposite half-spaces). In this case, to maximize the scalar product $a^Tu/\|u\|$ the vector $u$ will again try to point as close to $a$ as possible, but this time coming to the constraint $b^Tu=0$ (no normalization possible, $\|u_0\|=\infty$). Disregarding the constraint $b^Tu=1$, that is replacing it with non-strict $b^Tu\ge 0$, we will get that the best fit will be along the projection of $a$ onto the hyperplane $b^Tu=0$. It means that that is the maximum is not attained, only the supremum
$$
\sup=|\text{projection of $a$ onto $b^Tu=0$ along the normal $b$}|=
\sqrt{\|a\|^2-\frac{|a^Tb|^2}{\|b\|^2}}.
$$
A: Notice that your function is independent of the length of the $w$ vector. 
Since the given constraint merely normalizes $\|w\|$, the problem is effectively unconstrained.
Since the problem is unconstrained, there is no need to use the Lagrangian method.
Clear the radical by squaring of the function, then find the gradient 
$$\eqalign{
 \phi^2 &= \frac{w^TEE^Tw}{w^T\Sigma w} \cr
 \frac{\partial\phi}{\partial w} &= \bigg(\frac{(w^T\Sigma w)EE^T-(w^TEE^Tw)\Sigma }{(w^T\Sigma w)^2\phi}\bigg)\,w \cr
}$$
Setting this gradient to zero and clearing fractions yields
$$\eqalign{
\phi\,\Sigma\,w &= EE^Tw \cr
w &= \bigg(\frac{E^Tw}{\phi}\bigg)\,\Sigma^{-1}E = \lambda\,\Sigma^{-1}E \cr
}$$
Scale the result per the specified constraint
$$\eqalign{
w &= \frac{\lambda\Sigma^{-1}E}{i^T(\lambda\Sigma^{-1}E)}
  &= \frac{\Sigma^{-1}E}{i^T\Sigma^{-1}E}
}$$
