I have an optimization problem of the form $$\begin{align} \max_{\mathbf{x}} \quad \mathbf{x}^H \mathbf{A} \mathbf{x} \quad \text{s.t.} \quad \mathbf{b}^H\mathbf{x} = c \end{align}$$ where $\mathbf{b}$ is a vector of length $M$, $c$ is a scalar, and $\mathbf{A}$ is an invertible and definite positive matrix of $M \times M$. $^H$ is the complex conjugate transpose.
For this problem, the Lagrange multiplier method can be used, $$\begin{align} \mathcal{L}(\mathbf{x},\lambda) = \mathbf{x}^H \mathbf{A} \mathbf{x} + \lambda (\mathbf{b}^H\mathbf{x} - c) \end{align},$$ we have $$\begin{align} \frac{\partial \mathcal{L}(\mathbf{x},\lambda)}{\partial \mathbf{x}} &= 2\mathbf{A}\mathbf{x} + \lambda \mathbf{b} = \mathbf{0}, \\ \frac{\partial \mathcal{L}(\mathbf{x},\lambda)}{\partial \lambda} &= \mathbf{b} \mathbf{x} - c = 0 \end{align}.$$ The solution is $$\begin{align} \mathbf{x} = \frac{c \mathbf{A}^{-1}\mathbf{b}}{\mathbf{b}^H \mathbf{A}^{-1}\mathbf{b}} \end{align}.$$ which is the minimum. So, How to find its maximum?