I would like to find the vector $||\mathbf{x}||=1$ that maximizes the projection $\mathbf{s}^\top\mathbf{x}$ and minimizes $\mathbf{x}\mathbf{B}^{-1}\mathbf{x}^\top$, where $\mathbf{B}$ is a symmetric matrix.
I used Lagrange multipliers to set up the cost function
\begin{equation} J(\mathbf{x},\lambda)=-\mathbf{s}^\top\mathbf{x}+\mathbf{x}^\top\mathbf{B}^{-1}\mathbf{x}+\lambda(\mathbf{x}^\top\mathbf{x}-1) \end{equation}
and then tried to solve by setting $\frac{\partial J}{\partial\mathbf{x}}=\frac{\partial J}{\partial\lambda}=0$, yielding
\begin{eqnarray} \frac{\partial J}{\partial\mathbf{x}}&=2(\mathbf{B}^{-1}+\lambda\mathbf{I})\mathbf{x}-\mathbf{s}\\ &\implies2(\mathbf{B}^{-1}+\lambda\mathbf{I})\mathbf{x}=\mathbf{s} \end{eqnarray}
and
\begin{eqnarray} \frac{\partial J}{\partial\lambda}&=\mathbf{x}^\top\mathbf{x}-1\\ &\implies \mathbf{x}^\top\mathbf{x}=1. \end{eqnarray}
The trouble I'm having now is combining these two equations usefully. I can modify the first constraint to obtain
\begin{equation} \mathbf{x}=\frac{1}{2}(\mathbf{B}^{-1}+\lambda\mathbf{I})^{-1}\mathbf{s} \end{equation}
so that, taking the inner product and substituting the second constraint, I find
\begin{equation} \mathbf{x}^\top\mathbf{x}=1=\frac{1}{4}\mathbf{s}^\top(\mathbf{B}^{-1}+\lambda\mathbf{I})^{-T}(\mathbf{B}^{-1}+\lambda\mathbf{I})^{-1}\mathbf{s}. \end{equation}
At this point I'm not sure how to solve for $\lambda$. Any suggestions, or different procedures for solving this optimization problem, would be appreciated.