Corner solution of the maximization problem Answer










Hello, I upload the actual question with my 8-pages answer. Please can you check it. Is there a corner dissolution for $c=\gamma$. Please share your ideas. Thanks.
Here is the formulation of the problem :
\begin{eqnarray*} \max_{c, h, l} \  &amp; \ln (c - \gamma) + \beta l + \theta h \\ \text{s.t.} &amp;  l + h = 1, \\ &amp; c \leq \omega h + \rho, \\ \text{and} &amp; l, h \geq 0, c \geq \gamma \end{eqnarray*}

Substituting $l = 1 - h$, we can rewrite the above problem as :
\begin{eqnarray*} \max_{c, h} \  &amp; \ln (c - \gamma) + \beta + (\theta -\beta)h \\ \text{s.t.} &amp; 0 \leq h \leq 1 \\ \text{and} &amp; \gamma \leq c \leq \omega h + \rho\end{eqnarray*}

Since utility is increasing in $c$, $c = \omega h + \rho$ will hold in optimum. So we can further reduce the problem to :

\begin{eqnarray*} \max_{h} \  &amp; \ln (\omega h + \rho - \gamma) + \beta + (\theta -\beta)h \\ \text{s.t.} &amp; 0 \leq h \leq 1 \\ \text{and} &amp; \gamma \leq \omega h + \rho\end{eqnarray*}

Please note that we'll assume $\omega + \rho \geq \gamma$. This is because when $\omega + \rho &lt; \gamma$, there is no feasible solution. In other words, there does not exist any $h$ satisfying the constraints.

To solve this problem, we'll consider two cases :


Case 1: $\rho \geq \gamma$


In this case problem can be written as :

\begin{eqnarray*} \max_{h} \  &amp; \ln (\omega h + \rho - \gamma) + \beta + (\theta -\beta)h \\ \text{s.t.} &amp; 0 \leq h \leq 1 \end{eqnarray*}

Derivative of the objective with respect to $h$ is
$\frac{\omega}{\omega h + \rho - \gamma} + (\theta -\beta)$ which yields the following solution :

\begin{eqnarray*} h = \begin{cases} 1 &amp; \text{if } \frac{\omega}{\omega + \rho - \gamma} + (\theta -\beta) \geq 0  \\  0 &amp; \text{if } \frac{\omega}{\rho - \gamma} + (\theta -\beta) \leq 0  \\ \frac{1}{\beta -\theta} - \frac{\rho - \gamma}{\omega} &amp; \text{otherwise} \end{cases} \end{eqnarray*} 


Case 2: $\rho &lt; \gamma$


In this case problem can be written as :

\begin{eqnarray*} \max_{h}  \ &amp; \ln (\omega h + \rho - \gamma) + \beta + (\theta -\beta)h \\ \text{s.t.} &amp; \frac{\gamma - \rho}{\omega} \leq h \leq 1 \end{eqnarray*}

Derivative of the objective with respect to $h$ is
$\frac{\omega}{\omega h + \rho - \gamma} + (\theta -\beta)$ which yields the following solution :

\begin{eqnarray*} h = \begin{cases} 1 &amp; \text{if } \frac{\omega}{\omega + \rho - \gamma} + (\theta -\beta) \geq 0   \\ \frac{1}{\beta -\theta} - \frac{\rho - \gamma}{\omega} &amp; \text{otherwise} \end{cases} \end{eqnarray*}

Combining the two cases, we can write the solution as :

\begin{eqnarray*} h = \begin{cases} 1 &amp; \text{if } \frac{\omega}{\omega + \rho - \gamma} + (\theta -\beta) \geq 0  \\  0 &amp; \text{if } \frac{\omega}{\rho - \gamma} + (\theta -\beta) \leq 0 \text{ and } \rho \geq \gamma \\ \frac{1}{\beta -\theta} - \frac{\rho - \gamma}{\omega} &amp; \text{otherwise} \end{cases} \end{eqnarray*} 

Using $c = \omega h + \rho$ and $l = 1 -h$ we can get optimal values of $c$ and $l$ in each of the cases.