| \(w\): | \(h\): | \(\pi\): | \(T\): |
| Endogenous Variables and Parameters: | |||
| \(C\): | \(l\): | \(N_s\): | \(U\): |
Assuming that \(l < h\), we can set up the Lagrangian like so:
\[ \mathcal{L} = \ln(C) + \ln(l) - \lambda \cdot \left[ c - w\cdot(h-l) - \pi + T \right] \]Take the partial derivatives and set equal to \(0\) to get the first-order conditions:
\begin{align} 0 = \mathcal{L}_c^\prime &= \frac{1}{C} - \lambda \\ 0 = \mathcal{L}_l^\prime &= \frac{1}{l} - \lambda w \\ 0 = \mathcal{L}_\lambda^\prime &= - \left[ c - w\cdot(h-l) - \pi + T \right] \\ \end{align}Rearrange and combine to get:
\[\boxed{\begin{gather} w = {\left(\frac{1}{l}\right) \over \left(\frac{1}{C}\right)} = \frac{ C}{l} = MRS_{lC} \\ c = w\cdot(h-l) + \pi - T \\ \end{gather}}\]For this utility function, we can also explicitly solve for the optimal allocation. Rearrange to get \(C= w l \), then plug this into the budget constraint and solve for \(l\):
\begin{gather} w l = w\cdot(h-l) + \pi - T \\ 2lw = wh + \pi - T \\ l^* = \frac{1}{2}\frac{wh + \pi - T}{w}\\ l^* = \frac{1}{2}\left(h+\frac{ \pi - T}{w}\right)\\ \end{gather}And plug this into \(C=\frac{w}{\gamma} l \) to get:
\[C^* = \frac{wh + \pi - T}{2}\]Of course, if the above expressions give us an allocation in which \(l>h\), then our assumption that \(l < h \) was faulty, and the optimum is at the corner case, where \(l=h\) and \(C = \pi - T\).