<p>Hi game I see that quite well! I don't know if you have already seen Khun Tucker Conditions as you use a Lagrangian. because lagrangian is with equalities and Khun Tucker with inequalities, and as you see that's inequality and the matrix conditions change, but let's assume you didn't saw Khun Tucker and Lagrangian is okay, I find okey your answer, but that's because the Utility is convex too! And maybe having in consideration that the set is compact and close! you can show it in a graph.</p>

CC BY-SA 4.0
2018-05-12T21:29:51.047Z
Dear @MiguelGutierrez please can you show Kuhn Tucker solution? And can  you show what you said. This helps me very much. Thanks.
mnm123
CC BY-SA 4.0
2018-05-12T21:30:41.047Z
I have already shown the budget set is compact and closed. https://math.stackexchange.com/questions/2773486/a-set-which-is-closed-bounded-and-convex
mnm123
CC BY-SA 4.0
2018-05-12T21:30:55.980Z
But I cannot do what you say.
mnm123
CC BY-SA 4.0
2018-05-12T05:29:24.817Z
2018-05-12T05:29:24.817Z
Miguel Gutierrez
<p>Consider the following situation :
Preferences of the consumer are represented by the utility function $u(x_1, x_2) = x_1 + x_2$. Note that preferences are strictly monotonic and convex. Now let income be any $y &gt; 2$. So the consumer's problem is :
\begin{eqnarray*} \max_{x_1, x_2} &amp; x_1 + x_2 \\ \text{s.t.} &amp;  \ \ 2\sqrt{x_1} + x_2 \leq y \\ &amp; x_1\geq 0, \ x_2 \geq 0 \end{eqnarray*}
Now let us set up the Lagrangian :
\begin{eqnarray*} \mathcal{L}(x_1, x_2) = x_1+x_2 - \lambda(2\sqrt{x_1} + x_2 - y) +\mu_1x_1 + \mu_2x_2 \end{eqnarray*} 
Here are the first order necessary conditions:
\begin{eqnarray*} \dfrac{\partial\mathcal{L}}{\partial x_1} &amp; = &amp; 1 - \dfrac{\lambda}{\sqrt{x_1}} + \mu_1= 0 \\ \dfrac{\partial\mathcal{L}}{\partial x_2} &amp; = &amp; 1 - \lambda + \mu_2= 0 \\ \dfrac{\partial\mathcal{L}}{\partial \lambda} &amp; = &amp; y -2\sqrt{x_1} - x_2  \geq 0, \ \lambda\geq 0, \ \lambda(2\sqrt{x_1} + x_2 - y) = 0 \\ \dfrac{\partial\mathcal{L}}{\partial \mu_1} &amp; = &amp; x_1  \geq 0, \ \mu_1\geq 0, \ \mu_1x_1 = 0 \\ \dfrac{\partial\mathcal{L}}{\partial \mu_2} &amp; = &amp; x_2  \geq 0, \ \mu_2\geq 0, \ \mu_2x_2 = 0 \end{eqnarray*}</p>

<p>Check that one of the solutions to the above set of conditions is $x_1^* = 1, \ x_2^* = y -2, \ \lambda^*=1, \ \mu_1^* = 0, \ \mu_2^* = 0$, </p>

<p>But it is not the solution to the utility maximization problem stated above. This is because utility from $(x_1^*, x_2^*) = (1, y-2)$, i.e. $u(x_1^*, x_2^*) = 1 + y - 2= y-1$, is less than the utility from another affordable consumption bundle $(x^{**}_1, x^{**}_2) = (0, y)$ as $u(x_1^{**}, x_2^{**}) = 0 + y = y$ and $2\sqrt{x_1^{**}} + x_2^{**} =y$. </p>

<p>Therefore, first order conditions are not sufficient for utility maximization.</p>

CC BY-SA 4.0
2018-05-13T06:49:29.850Z
2018-05-13T06:49:29.850Z
Amit
<p>I have a budget set</p>
<p><span class="math-container">$$B=\{x=(x_1,x_2)\in R^2_+ \mid 2\sqrt{x_1}+x_2\le y\}$$</span></p>
<p>where <span class="math-container">$y&gt;0$</span> is income.</p>
<p>Assuming the preferences are strictly monotonic and convex, I want to show that first order conditions are necessary and sufficient for an interior solution to the utility maximization problem</p>
<p>**My solution **</p>
<p>First step : I maximize utility subject to the given budget set</p>
<p>Second step: I calculate first order conditions</p>
<p>Third step : I use hessian matrix.</p>
<p>If the determinant of this hessian matrix H is negative, then I can say that the FOCs are necessary is sufficient. (Is this true?)</p>
<p>My solution is like that <a href="https://i.sstatic.net/iFxvM.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/iFxvM.jpg" alt="enter image description here" /></a></p>
<p>where <span class="math-container">$$u_{11}={\partial^2 u(x_1, u_2)\over \partial x_1^2}$$</span></p>
<p><span class="math-container">$$u_{22}={\partial^2 u(x_1, u_2)\over \partial x_2^2}$$</span></p>
<p><span class="math-container">$$u_{12}={\partial^2 u(x_1, u_2)\over \partial x_1 \partial x_2}$$</span></p>
<p><span class="math-container">$$u_{21}={\partial^2 u(x_1, u_2)\over \partial x_1 \partial x_2}$$</span></p>
<p>I don’t know this answer is enough and correct. Because this solution seems not to be sufficient. Please share me your ideas about my solution.</p>
<p>Thanks a lot.</p>

<microeconomics><utility><consumer-theory><optimization><self-study>
Show that First order conditions are necessary and sufficient for utility maximization
mnm123
# Show that First order conditions are necessary and sufficient for utility maximization

(Asked by: username_0 on 2018-05-09)

I have a budget set

$$B=\{x=(x\_1,x\_2)\in R^2\_+ \mid 2\sqrt{x\_1}+x\_2\le y\}$$

where $y>0$ is income.

Assuming the preferences are strictly monotonic and convex, I want to show that first order conditions are necessary and sufficient for an interior solution to the utility maximization problem

\*\*My solution \*\*

First step : I maximize utility subject to the given budget set

Second step: I calculate first order conditions

Third step : I use hessian matrix.

If the determinant of this hessian matrix H is negative, then I can say that the FOCs are necessary is sufficient. (Is this true?)

My solution is like that [![enter image description here](https://i.sstatic.net/iFxvM.jpg)](https://i.sstatic.net/iFxvM.jpg)

where $$u\_{11}={\partial^2 u(x\_1, u\_2)\over \partial x\_1^2}$$

$$u\_{22}={\partial^2 u(x\_1, u\_2)\over \partial x\_2^2}$$

$$u\_{12}={\partial^2 u(x\_1, u\_2)\over \partial x\_1 \partial x\_2}$$

$$u\_{21}={\partial^2 u(x\_1, u\_2)\over \partial x\_1 \partial x\_2}$$

I don’t know this answer is enough and correct. Because this solution seems not to be sufficient. Please share me your ideas about my solution.

Thanks a lot.

**Tags:** microeconomics, utility, consumer-theory, optimization, self-study


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## 2 answers

### Answer by username_2 on 2018-05-13. Score: 3 | Accepted answer

Consider the following situation :
Preferences of the consumer are represented by the utility function $u(x\_1, x\_2) = x\_1 + x\_2$. Note that preferences are strictly monotonic and convex. Now let income be any $y > 2$. So the consumer's problem is :
\begin{eqnarray\*} \max\_{x\_1, x\_2} & x\_1 + x\_2 \\ \text{s.t.} & \ \ 2\sqrt{x\_1} + x\_2 \leq y \\ & x\_1\geq 0, \ x\_2 \geq 0 \end{eqnarray\*}
Now let us set up the Lagrangian :
\begin{eqnarray\*} \mathcal{L}(x\_1, x\_2) = x\_1+x\_2 - \lambda(2\sqrt{x\_1} + x\_2 - y) +\mu\_1x\_1 + \mu\_2x\_2 \end{eqnarray\*}
Here are the first order necessary conditions:
\begin{eqnarray\*} \dfrac{\partial\mathcal{L}}{\partial x\_1} & = & 1 - \dfrac{\lambda}{\sqrt{x\_1}} + \mu\_1= 0 \\ \dfrac{\partial\mathcal{L}}{\partial x\_2} & = & 1 - \lambda + \mu\_2= 0 \\ \dfrac{\partial\mathcal{L}}{\partial \lambda} & = & y -2\sqrt{x\_1} - x\_2 \geq 0, \ \lambda\geq 0, \ \lambda(2\sqrt{x\_1} + x\_2 - y) = 0 \\ \dfrac{\partial\mathcal{L}}{\partial \mu\_1} & = & x\_1 \geq 0, \ \mu\_1\geq 0, \ \mu\_1x\_1 = 0 \\ \dfrac{\partial\mathcal{L}}{\partial \mu\_2} & = & x\_2 \geq 0, \ \mu\_2\geq 0, \ \mu\_2x\_2 = 0 \end{eqnarray\*}

Check that one of the solutions to the above set of conditions is $x\_1^\* = 1, \ x\_2^\* = y -2, \ \lambda^\*=1, \ \mu\_1^\* = 0, \ \mu\_2^\* = 0$,

But it is not the solution to the utility maximization problem stated above. This is because utility from $(x\_1^\*, x\_2^\*) = (1, y-2)$, i.e. $u(x\_1^\*, x\_2^\*) = 1 + y - 2= y-1$, is less than the utility from another affordable consumption bundle $(x^{\*\*}\_1, x^{\*\*}\_2) = (0, y)$ as $u(x\_1^{\*\*}, x\_2^{\*\*}) = 0 + y = y$ and $2\sqrt{x\_1^{\*\*}} + x\_2^{\*\*} =y$.

Therefore, first order conditions are not sufficient for utility maximization.


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### Answer by username_1 on 2018-05-12. Score: 0

Hi game I see that quite well! I don't know if you have already seen Khun Tucker Conditions as you use a Lagrangian. because lagrangian is with equalities and Khun Tucker with inequalities, and as you see that's inequality and the matrix conditions change, but let's assume you didn't saw Khun Tucker and Lagrangian is okay, I find okey your answer, but that's because the Utility is convex too! And maybe having in consideration that the set is compact and close! you can show it in a graph.

> **Comments:**
> **username_0**: Dear @username_1 please can you show Kuhn Tucker solution? And can  you show what you said. This helps me very much. Thanks.  
> **username_0**: I have already shown the budget set is compact and closed. https://math.stackexchange.com/questions/2773486/a-set-which-is-closed-bounded-and-convex  
> **username_0**: But I cannot do what you say.  

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