# Corner solution of the maximization problem 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.

• – Rodrigo de Azevedo Oct 23 '19 at 9:33
• How do you know there's a corner solution? – Art Oct 23 '19 at 15:42
• @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? – B11b Oct 23 '19 at 15:46
• Don´t we have here an equation as a constraint, $h+l=T$? – callculus Oct 23 '19 at 16:36
• 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 Oct 23 '19 at 20:55

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.

• I donu Knopf how to thank you!! You are the best and your solution is so clever and perfect !! Thanks again Amit – B11b Nov 8 '19 at 23:49

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.

• 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? – B11b Oct 23 '19 at 19:05
• @user315 " Can I upload my solution by hand-writing? " Yes, you can do that by editing your question. – callculus Oct 23 '19 at 19:11
• Thanks a lot... – B11b Oct 23 '19 at 19:13
• @callculus I uploaded. Please check it. And please tell me whether it is correct or not. Thanks a lot. – B11b Oct 23 '19 at 19:28
• @Regio I added my solution. – B11b Oct 23 '19 at 19:28