<p>The corner solution is not <span class="math-container">$c=a$</span> 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 <span class="math-container">$h=0$</span>. Since the agent has non-labor income <span class="math-container">$p$</span>, 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. </p>

<p>After solving for the optimal <span class="math-container">$l$</span>, <span class="math-container">$c$</span> and <span class="math-container">$h$</span> I am sure that the optimal <span class="math-container">$h$</span> is defined by the  difference between two terms. Since you cannot work negative hours, the corner solution occurs whenever your equation for <span class="math-container">$h$</span> goes negative. </p>

<p>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. </p>

CC BY-SA 4.0
2019-10-23T19:05:09.080Z
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?
studentp
CC BY-SA 4.0
2019-10-23T19:11:33.860Z
@user315 " *Can I upload my solution by hand-writing?* " Yes, you can do that by editing your question.
callculus42
CC BY-SA 4.0
2019-10-23T19:28:25.630Z
@callculus I uploaded. Please check it. And please tell me whether it is correct or not. Thanks a lot.
studentp
CC BY-SA 4.0
2019-10-23T19:28:47.327Z
@Regio I added my solution.
studentp
CC BY-SA 4.0
2019-10-23T20:54:27.113Z
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.
Regio
CC BY-SA 4.0
2019-10-23T19:01:57.700Z
2019-10-23T19:01:57.700Z
Regio
<p>Here is the formulation of the problem :
<span class="math-container">\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*}</span></p>

<p>Substituting <span class="math-container">$l = 1 - h$</span>, we can rewrite the above problem as :
<span class="math-container">\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*}</span></p>

<p>Since utility is increasing in <span class="math-container">$c$</span>, <span class="math-container">$c = \omega h + \rho$</span> will hold in optimum. So we can further reduce the problem to :</p>

<p><span class="math-container">\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*}</span></p>

<p>Please note that we'll assume <span class="math-container">$\omega + \rho \geq \gamma$</span>. This is because when <span class="math-container">$\omega + \rho &lt; \gamma$</span>, there is no feasible solution. In other words, there does not exist any <span class="math-container">$h$</span> satisfying the constraints.</p>

<p>To solve this problem, we'll consider two cases :</p>

<ul>
<li><strong>Case 1</strong>: <span class="math-container">$\rho \geq \gamma$</span></li>
</ul>

<p>In this case problem can be written as :</p>

<p><span class="math-container">\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*}</span></p>

<p>Derivative of the objective with respect to <span class="math-container">$h$</span> is
<span class="math-container">$\frac{\omega}{\omega h + \rho - \gamma} + (\theta -\beta)$</span> which yields the following solution :</p>

<p><span class="math-container">\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*}</span> </p>

<ul>
<li><strong>Case 2</strong>: <span class="math-container">$\rho &lt; \gamma$</span></li>
</ul>

<p>In this case problem can be written as :</p>

<p><span class="math-container">\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*}</span></p>

<p>Derivative of the objective with respect to <span class="math-container">$h$</span> is
<span class="math-container">$\frac{\omega}{\omega h + \rho - \gamma} + (\theta -\beta)$</span> which yields the following solution :</p>

<p><span class="math-container">\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*}</span></p>

<p>Combining the two cases, we can write the solution as :</p>

<p><span class="math-container">\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*}</span> </p>

<p>Using <span class="math-container">$c = \omega h + \rho$</span> and <span class="math-container">$l = 1 -h$</span> we can get optimal values of <span class="math-container">$c$</span> and <span class="math-container">$l$</span> in each of the cases.</p>

CC BY-SA 4.0
2019-11-08T23:49:36.193Z
I donu Knopf how to thank you!! You are the best and your solution is so clever and perfect !! Thanks again Amit
studentp
CC BY-SA 4.0
2019-10-24T01:49:08.203Z
2019-11-09T00:37:24.950Z
2019-11-09T00:37:24.950Z
Amit
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<p>Answer</p>

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<p>Hello, I upload the actual question with my 8-pages answer. Please can you check it. Is there a corner dissolution for <span class="math-container">$c=\gamma$</span>. Please share your ideas. Thanks. </p>

CC BY-SA 4.0
2019-10-23T09:33:41.073Z
X-posted: https://math.stackexchange.com/q/3405439/339790
Rodrigo de Azevedo
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2019-10-23T15:42:11.943Z
How do you know there's a corner solution?
Art
CC BY-SA 4.0
2019-10-23T15:46:44.777Z
@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?
studentp
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2019-10-23T16:36:51.530Z
Don´t we have here an equation as a constraint, $h+l=T$?
callculus42
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2019-10-23T17:20:03.403Z
Yes I put two constraints. But also it has a solution for $c=a$. But how? @callculus
studentp
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2019-10-23T17:22:17.310Z
"Yes I put two constraints" What does that mean?
callculus42
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2019-10-23T17:25:49.407Z
The constraint $l+h\le T$ that you said is time constraint. And also I use budget constraint. And I solve this optimization problem by subject to these constraint. But my question (what I could not do) is different from what you said @callculus
studentp
CC BY-SA 4.0
2019-10-23T20:55:31.773Z
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."
BKay
<consumer-theory><self-study><optimization><linear-programming>
Corner solution of the maximization problem
studentp
# Corner solution of the maximization problem

(Asked by: username_0 on 2019-10-23)

[![Question](https://i.sstatic.net/TvSwV.jpg)](https://i.sstatic.net/TvSwV.jpg)

Answer

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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.

**Tags:** consumer-theory, self-study, optimization, linear-programming
> **Comments:**
> **username_1**: X-posted: https://math.stackexchange.com/q/3405439/339790  
> **username_2**: How do you know there's a corner solution?  
> **username_0**: @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?  
> **username_3**: Don´t we have here an equation as a constraint, $h+l=T$?  
> **username_0**: Yes I put two constraints. But also it has a solution for $c=a$. But how? @callculus  
> **username_3**: "Yes I put two constraints" What does that mean?  
> **username_0**: The constraint $l+h\le T$ that you said is time constraint. And also I use budget constraint. And I solve this optimization problem by subject to these constraint. But my question (what I could not do) is different from what you said @callculus  
> **username_4**: 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."  

---

## 2 answers

### Answer by username_6 on 2019-10-24. Score: 3 | Accepted answer

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.

> **Comments:**
> **username_0**: I donu Knopf how to thank you!! You are the best and your solution is so clever and perfect !! Thanks again Amit  

---

### Answer by username_5 on 2019-10-23. Score: 1

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.

> **Comments:**
> **username_0**: 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?  
> **username_3**: @user315 " *Can I upload my solution by hand-writing?* " Yes, you can do that by editing your question.  
> **username_0**: @callculus I uploaded. Please check it. And please tell me whether it is correct or not. Thanks a lot.  
> **username_0**: @Regio I added my solution.  
> **username_5**: 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.  

---

