$\mu$ and $\lambda$ are two positive constants and $\mu > \lambda$. $c$ and $b$ are two positive constants and $c < b$. How to solve the following optimization problem?
\begin{align} \max_{T\ge 0, H_t} & \, \frac{W_0}{M_0} \\ s.t. & \, \frac{dW_t}{dt} = \mu W_t - \mu H_t, H_t \ge W_t + c, W_T = 0. \\ & \frac{dM_t}{dt} = \lambda M_t - \lambda H_t, H_t \ge M_t + b, M_T = 0. \end{align}
Using a state space formulation with the state vector, $x_t$, as $\begin{bmatrix} W_t \\ M_t \end{bmatrix}$ and $H_t$ as the input, $u_t$, $$\begin{bmatrix} \dot{W_t} \\ \dot{M_t} \\ \end{bmatrix} = \begin{bmatrix} \mu & 0 \\ 0 & \lambda \end{bmatrix} \begin{bmatrix} W_t \\ M_t \end{bmatrix} + \begin{bmatrix} -\mu \\ -\lambda \end{bmatrix} H_t \\ \dot{x}_t = Ax_t + Bu_t$$
Taking its Laplace transform,
$$sX(s) - x_0 = AX(s) + BU(s) \\ X(s)(sI - A) = BU(s) + x_0 \\ X(s) = (sI - A)^{-1}(BU(s) + x_0) \\ x_t = \mathcal{L}^{-1}[(sI-A)^{-1}(BU(s) + x_0)] \\ \begin{bmatrix} W_t \\ M_t \end{bmatrix} = \begin{bmatrix} -\mu e^{\mu t}* H_t + e^{\mu t}W_0 \\ -\lambda e^{\lambda t}*H_t + e^{\lambda t}M_0 \end{bmatrix}$$
A little algebra gives us $$W_0 = (W_t + \mu e^{\mu t} * H_t )e^{-\mu t}\\ M_0 = (M_t + \lambda e^{\lambda t} * H_t )e^{-\lambda t} $$ Using $W_T = M_T = 0$, we get $$W_0 = (\mu e^{\mu T} * H_T) e^{-\mu T} \\ M_0 = (\lambda e^{\lambda T} * H_T) e^{-\lambda T}$$
Since $e^{\mu T}, e^{\lambda T}$ and $H_T$ are constants for a particular $T$, their convolution w.r.t time yields $te^{\mu T}H_T $ and $te^{\lambda t}H_T $. Therefore, the objective function becomes $\frac{W_0}{M_0} = \frac{\mu}{\lambda}$, which is independent of $T, H_t$.