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Hmm, you could rewrite it this way: (I'm going to assume that $\pi_0, \pi_1, \xi'', \mu, \lambda$ are pre-defined constants, and $\alpha,W$ are variables)

$$ \begin{align} &\max_{W}\; \pi_{0} (1 - e^{-\mu W}) - \frac{\pi_{1}}{W}z_{1}(W)\\ s.t.\;& \frac{dz_{0}(\alpha)}{d\alpha} = 1 - e^{-\mu \alpha},\quad z_{0}(0) = 0\\ & \frac{dz_{1}(\alpha)}{d\alpha} = 1 - e^{-\lambda \alpha},\quad z_{1}(0) = 0\\ & z_{0}(W) + \epsilon \leq \xi''\\ & W \geq 0 \end{align} $$ where $z_{0},z_{1}$ are auxiliary variables and $\epsilon$ is a numerical tolerance value. This then becomes a DAE (differential-algebraic equation) optimization problem, which can be solved numerically. (though given that your decision variable is also the independent variable in the differential equations, some further bilinear transformations may be required).

Edit: I just realized, if indeed $\xi''$ is a constant as I have assumed, the inequality constraint can easily be converted into a bound. $$ \begin{align} \int_{0}^{W^U} 1-e^{-\mu \alpha}\,d\alpha = \xi'' - \epsilon\\ \frac{e^{-\mu W^{U}}}{\mu} + W^{U} - \frac{1}{\mu} = \xi'' - \epsilon\\ \end{align} $$ Solve for $W^{U}$, and replace the above inequality constraints with: $$ 0 \leq W \leq W^{U} $$ You may be able to solve this using optimal control methods.

Hmm, you could rewrite it this way: (I'm going to assume that $\pi_0, \pi_1, \xi'', \mu, \lambda$ are pre-defined constants, and $\alpha,W$ are variables)

$$ \begin{align} &\max_{W}\; \pi_{0} (1 - e^{-\mu W}) - \frac{\pi_{1}}{W}z_{1}(W)\\ s.t.\;& \frac{dz_{0}(\alpha)}{d\alpha} = 1 - e^{-\mu \alpha},\quad z_{0}(0) = 0\\ & \frac{dz_{1}(\alpha)}{d\alpha} = 1 - e^{-\lambda \alpha},\quad z_{1}(0) = 0\\ & z_{0}(W) + \epsilon \leq \xi''\\ & W \geq 0 \end{align} $$ where $z_{0},z_{1}$ are auxiliary variables and $\epsilon$ is a numerical tolerance value. This then becomes a DAE (differential-algebraic equation) optimization problem, which can be solved numerically. (though given that your decision variable is also the independent variable in the differential equations, some further bilinear transformations may be required. See http://dx.doi.org/10.1016/j.na.2005.03.066. Some software packages do this automatically.).

Edit: I just realized, if indeed $\xi''$ is a constant as I have assumed, the inequality constraint can easily be converted into a bound. $$ \begin{align} \int_{0}^{W^U} 1-e^{-\mu \alpha}\,d\alpha = \xi'' - \epsilon\\ \frac{e^{-\mu W^{U}}}{\mu} + W^{U} - \frac{1}{\mu} = \xi'' - \epsilon\\ \end{align} $$ Solve for $W^{U}$, and replace the above inequality constraints with: $$ 0 \leq W \leq W^{U} $$ You may be able to solve this using optimal control methods.

Hmm, you could rewrite it this way: (I'm going to assume that $\pi_0, \pi_1, \xi'', \mu, \lambda$ are pre-defined constants, and $\alpha,W$ are variables)

$$ \begin{align} &\max_{W}\; \pi_{0} (1 - e^{-\mu W}) - \frac{\pi_{1}}{W}z_{1}(W)\\ s.t.\;& \frac{dz_{0}(\alpha)}{d\alpha} = 1 - e^{-\mu \alpha},\quad z_{0}(0) = 0\\ & \frac{dz_{1}(\alpha)}{d\alpha} = 1 - e^{-\lambda \alpha},\quad z_{1}(0) = 0\\ & z_{0}(W) + \epsilon \leq \xi''\\ & W \geq 0 \end{align} $$ where $z_{0},z_{1}$ are auxiliary variables and $\epsilon$ is a numerical tolerance value. This then becomes a DAE (differential-algebraic equation) optimization problem.

Edit: I just realized, if indeed $\xi''$ is a constant as I have assumed, the inequality constraint can easily be converted into a bound. $$ \begin{align} \int_{0}^{W^U} 1-e^{-\mu \alpha}\,d\alpha = \xi'' - \epsilon\\ \frac{e^{-\mu W^{U}}}{\mu} + W^{U} - \frac{1}{\mu} = \xi'' - \epsilon\\ \end{align} $$ Solve for $W^{U}$, and replace the above inequality constraints with: $$ 0 \leq W \leq W^{U} $$

Hmm, you could rewrite it this way: (I'm going to assume that $\pi_0, \pi_1, \xi'', \mu, \lambda$ are pre-defined constants, and $\alpha,W$ are variables)

$$ \begin{align} &\max_{W}\; \pi_{0} (1 - e^{-\mu W}) - \frac{\pi_{1}}{W}z_{1}(W)\\ s.t.\;& \frac{dz_{0}(\alpha)}{d\alpha} = 1 - e^{-\mu \alpha},\quad z_{0}(0) = 0\\ & \frac{dz_{1}(\alpha)}{d\alpha} = 1 - e^{-\lambda \alpha},\quad z_{1}(0) = 0\\ & z_{0}(W) + \epsilon \leq \xi''\\ & W \geq 0 \end{align} $$ where $z_{0},z_{1}$ are auxiliary variables and $\epsilon$ is a numerical tolerance value. This then becomes a DAE (differential-algebraic equation) optimization problem, which can be solved numerically. (though given that your decision variable is also the independent variable in the differential equations, some further bilinear transformations may be required).

Edit: I just realized, if indeed $\xi''$ is a constant as I have assumed, the inequality constraint can easily be converted into a bound. $$ \begin{align} \int_{0}^{W^U} 1-e^{-\mu \alpha}\,d\alpha = \xi'' - \epsilon\\ \frac{e^{-\mu W^{U}}}{\mu} + W^{U} - \frac{1}{\mu} = \xi'' - \epsilon\\ \end{align} $$ Solve for $W^{U}$, and replace the above inequality constraints with: $$ 0 \leq W \leq W^{U} $$ You may be able to solve this using optimal control methods.

Hmm, you could rewrite it this way: (I'm going to assume that $\pi_0, \pi_1, \xi'', \mu, \lambda$ are pre-defined constants, and $\alpha,W$ are variables)

$$ \begin{align} &\max_{W}\; \pi_{0} (1 - e^{-\mu W}) - \frac{\pi_{1}}{W}z_{1}(W)\\ s.t.\;& \frac{dz_{0}(\alpha)}{d\alpha} = 1 - e^{-\mu \alpha},\quad z_{0}(0) = 0\\ & \frac{dz_{1}(\alpha)}{d\alpha} = 1 - e^{-\lambda \alpha},\quad z_{1}(0) = 0\\ & z_{0}(W) + \epsilon \leq \xi''\\ & W \geq 0 \end{align} $$ where $z_{0},z_{1}$ are auxiliary variables and $\epsilon$ is a numerical tolerance value. This then becomes a DAE (differential-algebraic equation) optimization problem. I don't believe there exists an analytical solution for this problem, owing to the algebraic inequality constraint (otherwise one could apply Pontrayagin's maximum principle) -- but it is certainly possible to solve this system numerically.

Mind you, this is a nonconvex optimization problem, so a global solution is not guaranteed if you apply local methods.

Hmm, you could rewrite it this way: (I'm going to assume that $\pi_0, \pi_1, \xi'', \mu, \lambda$ are pre-defined constants, and $\alpha,W$ are variables)

$$ \begin{align} &\max_{W}\; \pi_{0} (1 - e^{-\mu W}) - \frac{\pi_{1}}{W}z_{1}(W)\\ s.t.\;& \frac{dz_{0}(\alpha)}{d\alpha} = 1 - e^{-\mu \alpha},\quad z_{0}(0) = 0\\ & \frac{dz_{1}(\alpha)}{d\alpha} = 1 - e^{-\lambda \alpha},\quad z_{1}(0) = 0\\ & z_{0}(W) + \epsilon \leq \xi''\\ & W \geq 0 \end{align} $$ where $z_{0},z_{1}$ are auxiliary variables and $\epsilon$ is a numerical tolerance value. This then becomes a DAE (differential-algebraic equation) optimization problem.

Edit: I just realized, if indeed $\xi''$ is a constant as I have assumed, the inequality constraint can easily be converted into a bound. $$ \begin{align} \int_{0}^{W^U} 1-e^{-\mu \alpha}\,d\alpha = \xi'' - \epsilon\\ \frac{e^{-\mu W^{U}}}{\mu} + W^{U} - \frac{1}{\mu} = \xi'' - \epsilon\\ \end{align} $$ Solve for $W^{U}$, and replace the above inequality constraints with: $$ 0 \leq W \leq W^{U} $$