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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.

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    $\begingroup$ X-posted: math.stackexchange.com/q/3405439/339790 $\endgroup$ Oct 23, 2019 at 9:33
  • $\begingroup$ How do you know there's a corner solution? $\endgroup$
    – Art
    Oct 23, 2019 at 15:42
  • $\begingroup$ @Art nothing. I do just solve its interior solutions. But I have learnt that I need to find its corner solutions as well. But I don’t know (no idea) about how to find corner solution. Please can you help me? $\endgroup$
    – studentp
    Oct 23, 2019 at 15:46
  • $\begingroup$ Don´t we have here an equation as a constraint, $h+l=T$? $\endgroup$ Oct 23, 2019 at 16:36
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    $\begingroup$ This is potentially a great question but currently I'm voting to close this question as off-topic because it fails to follow site policy on home work: "Do not merely post a scan or image of the whole question, nor of your attempted answer. Enter your question, and the work you've done to try to answer it, as text." $\endgroup$
    – BKay
    Oct 23, 2019 at 20:55

2 Answers 2

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Here is the formulation of the problem : \begin{eqnarray*} \max_{c, h, l} \ & \ln (c - \gamma) + \beta l + \theta h \\ \text{s.t.} & l + h = 1, \\ & c \leq \omega h + \rho, \\ \text{and} & 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} \ & \ln (c - \gamma) + \beta + (\theta -\beta)h \\ \text{s.t.} & 0 \leq h \leq 1 \\ \text{and} & \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} \ & \ln (\omega h + \rho - \gamma) + \beta + (\theta -\beta)h \\ \text{s.t.} & 0 \leq h \leq 1 \\ \text{and} & \gamma \leq \omega h + \rho\end{eqnarray*}

Please note that we'll assume $\omega + \rho \geq \gamma$. This is because when $\omega + \rho < \gamma$, 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} \ & \ln (\omega h + \rho - \gamma) + \beta + (\theta -\beta)h \\ \text{s.t.} & 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 & \text{if } \frac{\omega}{\omega + \rho - \gamma} + (\theta -\beta) \geq 0 \\ 0 & \text{if } \frac{\omega}{\rho - \gamma} + (\theta -\beta) \leq 0 \\ \frac{1}{\beta -\theta} - \frac{\rho - \gamma}{\omega} & \text{otherwise} \end{cases} \end{eqnarray*}

  • Case 2: $\rho < \gamma$

In this case problem can be written as :

\begin{eqnarray*} \max_{h} \ & \ln (\omega h + \rho - \gamma) + \beta + (\theta -\beta)h \\ \text{s.t.} & \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 & \text{if } \frac{\omega}{\omega + \rho - \gamma} + (\theta -\beta) \geq 0 \\ \frac{1}{\beta -\theta} - \frac{\rho - \gamma}{\omega} & \text{otherwise} \end{cases} \end{eqnarray*}

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

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

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    $\begingroup$ I donu Knopf how to thank you!! You are the best and your solution is so clever and perfect !! Thanks again Amit $\endgroup$
    – studentp
    Nov 8, 2019 at 23:49
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The corner solution is not $c=a$ it cannot be because the marginal utility of even a tiny bit of consumption is unbounded there. However, you can have a corner solution where $h=0$. Since the agent has non-labor income $p$, the budget line has a kink. That is, if the agent receives a lot of income even without working, they may choose not to work, and fully enjoy leisure.

After solving for the optimal $l$, $c$ and $h$ I am sure that the optimal $h$ is defined by the difference between two terms. Since you cannot work negative hours, the corner solution occurs whenever your equation for $h$ goes negative.

If you are not sure what I mean, simply update your question with the actual formulas you obtained, I can provide further comments and guide you to find the corner solution.

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  • $\begingroup$ Yes, I could not find for $c=a$. But, someone says there exist. Can I upload my solution by hand-writing? Because solution is too long and my writing is very readable and good. Do you accept dear Regio? $\endgroup$
    – studentp
    Oct 23, 2019 at 19:05
  • $\begingroup$ @user315 " Can I upload my solution by hand-writing? " Yes, you can do that by editing your question. $\endgroup$ Oct 23, 2019 at 19:11
  • $\begingroup$ @callculus I uploaded. Please check it. And please tell me whether it is correct or not. Thanks a lot. $\endgroup$
    – studentp
    Oct 23, 2019 at 19:28
  • $\begingroup$ @Regio I added my solution. $\endgroup$
    – studentp
    Oct 23, 2019 at 19:28
  • $\begingroup$ The solution looks fine to me. I see now that there can be corner solution if $p+hw=\gamma$, in that case, even if you work as much as possible, you can only afford $\gamma$ units of consumption. $\endgroup$
    – Regio
    Oct 23, 2019 at 20:54

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