# Consumer's Problem

Exogenous Variables and Parameters:
 $$w$$: $$h$$: $$\pi$$: $$T$$: Endogenous Variables and Parameters: $$C$$: test $$l$$: test $$N_s$$: test $$U$$: test

### Solving the problem algebraically:

The consumer's problem is $\max_{C,l}\; U(C,l)$ subject to the constraints: \begin{gather} c\geq 0, \quad \quad h \geq l \geq 0 \tag{NonNeg}\\ c \leq w\cdot(h-l) + \pi - T \tag{Budget}\\ \end{gather}

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:

$\boxed{\begin{gather} w = {\left(\frac{1}{l}\right) \over \left(\frac{1}{C}\right)} = \frac{ C}{l} = MRS_{lC} \\ c = w\cdot(h-l) + \pi - T \\ \end{gather}}$
###### Explicitly solving:

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 $$C = \pi - T$$.