Let $C$ a positive definite matrix, $\beta$ and $\epsilon$ constant parameters, $\alpha$, $\eta$, decision variables.
$$J = \textrm{Tr}\left(\begin{bmatrix}\alpha & -2\beta\eta \\ \alpha & (1 - 2\beta \eta)\end{bmatrix}'C \begin{bmatrix}\alpha & -2\beta\eta \\ \alpha & (1 - 2\beta \eta)\end{bmatrix} + \begin{bmatrix}\eta \\ \eta\end{bmatrix}\epsilon^2\begin{bmatrix}\eta & \eta\end{bmatrix} \right) $$
I know $J\ge0$ because the trace of the sum of two quadratic forms, how to find the optimal solution?
I was expecting to be able to cast this problem to the standard form
$$\begin{bmatrix}\alpha \\ \eta \end{bmatrix}' A \begin{bmatrix}\alpha \\ \eta \end{bmatrix} + b \begin{bmatrix}\alpha \\ \eta \end{bmatrix} + c $$
Edit:
I was asked in the comments for what I can get with a CAS. Evaluating $J$ then extracting the coefficients I can get a solution:
$$\displaystyle \left[\begin{matrix}\alpha\\\eta\end{matrix}\right]^{T} \left[\begin{matrix}2 {c}_{0,0} & - 4 \beta {c}_{0,1}\\- 4 \beta {c}_{0,1} & 2 \left(4 \beta^{2} {c}_{1,1} + \epsilon^{2}\right)\end{matrix}\right] \left[\begin{matrix}\alpha\\\eta\end{matrix}\right] + \left[\begin{matrix}2 {c}_{0,1} & - 4 \beta {c}_{1,1}\end{matrix}\right] \left[\begin{matrix}\alpha\\\eta\end{matrix}\right] + \left[\begin{matrix}{c}_{1,1}\end{matrix}\right]$$
But I wanted to see if someone could point something less tedious
