<p>For part (i), in complete rigor, you should also check <a href="https://mjo.osborne.economics.utoronto.ca/index.php/tutorial/index/1/qcc/t#d:borderedHessian" rel="nofollow noreferrer">the determinants of all the leading principal minors of the bordered Hessian and make sure that they have alternating signs</a>. Your final conclusion looks correct though.</p>
<p>For part (ii), recall that Walrasian demand is the solution to utility maximization subject to budget constraint. So you should setup a Lagrangian, derive the <a href="https://mjo.osborne.economics.utoronto.ca/index.php/tutorial/index/1/ktc/t" rel="nofollow noreferrer">Kuhn-Tucker conditions</a>, and then solve for <span class="math-container">$x,y,z$</span> as functions of the prices and income. These will be the Walrasian demands.</p>

CC BY-SA 4.0
2021-11-17T11:27:05.763Z
Thanks for your response. I have updated my question with the Kuhn Tucker conditions. Because z, disappears I am unable to isolate x, y, and z just in terms of prices and wealth, which is what the Walrasian correspondence demands.
Kinno
CC BY-SA 4.0
2021-11-17T13:39:45.337Z
@Kinno: You can usually recover $z$ from the budget constraint after solving for $x$ and $y$. You should probably start from your condition $3$, and use the fact/assumption that $p_1,p_2,p_3>0$ to deduce that $\lambda>0$, which in turn will give you strict equality in condition $7$, and go from there.
Herr K.
CC BY-SA 4.0
2021-11-17T13:48:46.917Z
@Kinno: Also, the MU-to-Price Ratio gives you a relation between $x$ and $y$, e.g. $x=2yp_2/p_1$. Plug this into either $1$ or $2$ for $x$, and you'll get $y$ as an expression of prices and $\lambda$.
Herr K.
CC BY-SA 4.0
2021-11-24T23:07:50.893Z
When all are positive x,y,z>0 there is an infinity of solutions. This only happens at particular prices. Walras demand is in this knife-edge case not strictly speaking a function of price and income as suggested by @Herr K.
Jesper Hybel
CC BY-SA 4.0
2021-11-25T01:55:21.537Z
@JesperHybel: Yes you're right. Thanks for pointing that out.
Herr K.
CC BY-SA 4.0
2021-11-16T20:54:13.233Z
2021-11-16T20:54:13.233Z
Herr K.
<p><span class="math-container">$u(x,y,z)=x^{\frac{2}{3}}y^{\frac{1}{3}}+z$</span> is the sum of two concave functions: <span class="math-container">$u_1(x,y,z)=x^{\frac{2}{3}}y^{\frac{1}{3}}$</span> and <span class="math-container">$u_2(x,y,z)=z$</span>, and therefore, it is concave. Since <span class="math-container">$u$</span> is concave, it is also quasi-concave. Preferences represented by <span class="math-container">$u$</span> are, therefore, convex. However, these preferences are not strictly convex. To show this, consider <span class="math-container">$(x_1,y_1,z_1)=(2,2,0)$</span> and <span class="math-container">$(x_2,y_2,z_2)=(0,0,2)$</span>, and <span class="math-container">$\lambda = \frac{1}{2}$</span>. Clearly,
<span class="math-container">$u(\lambda (x_1,y_1,z_1) + (1-\lambda)(x_2,y_2,z_2))=u(1,1,1)=2=u(2,2,0)=u(0,0,2)=\min(u(x_1,y_1,z_1),u(x_2,y_2,z_2))$</span></p>
<p>Demand can be obtained by solving this problem:
<span class="math-container">\begin{eqnarray*} \max_{x,y,z} &amp; \ x^{\frac{2}{3}}y^{\frac{1}{3}}+z \\ \text{s.t. } &amp; p_Xx+p_Yy+p_Zz\leq M \\ \text{and } &amp; x\geq 0, \ y\geq 0, \ z \geq 0\end{eqnarray*}</span>
where <span class="math-container">$p_X&gt;0, p_Y&gt;0, p_Z&gt;0, M\geq 0$</span></p>
<p>and we get the demand as</p>
<p><span class="math-container">$(x^d,y^d,z^d)(p_X,p_Y,p_Z,M)\in\begin{cases} \left\{\left(\dfrac{2M}{3p_X},\dfrac{M}{3p_Y},0\right)\right\} &amp;\text{if } \frac{3p_X^{\frac{2}{3}}p_Y^{\frac{1}{3}}}{2^{\frac{2}{3}}}&lt;p_Z \\ \left\{\left(0,0,\dfrac{M}{p_Z}\right)\right\} &amp; \text{if } \frac{3p_X^{\frac{2}{3}}p_Y^{\frac{1}{3}}}{2^{\frac{2}{3}}}&gt;p_Z \\ \left\{\left(\dfrac{2\alpha M}{3p_X},\dfrac{\alpha M}{3p_Y},\dfrac{(1-\alpha)M}{p_Z}\right)|0\leq\alpha\leq 1\right\}&amp;\text{if } \frac{3p_X^{\frac{2}{3}}p_Y^{\frac{1}{3}}}{2^{\frac{2}{3}}}=p_Z\end{cases}$</span></p>
<p>Note: Approach I have used to find the demand here is same as the one I have used here: <a href="https://economics.stackexchange.com/a/51562/11824">https://economics.stackexchange.com/a/51562/11824</a></p>

CC BY-SA 4.0
2023-10-15T19:38:25.150Z
2023-10-16T00:15:11.220Z
2023-10-16T00:15:11.220Z
Amit
<p>I am given <span class="math-container">$U(x,y,z) = x^\frac{2}{3}y^\frac{1}{3} + z$</span>. I am asked to solve the following:</p>
<p>(i) Prove the convexity of these preferences (convex, strictly convex or neither?)<br />
(ii) Solve for the Walrasian Demand?</p>
<p>For part 1, I calculated the determinant of the bordered hessian matrix and got  <span class="math-container">$\frac{320}{81}*\frac{1}{x^\frac{2}{3}y^\frac{4}{3}}$</span>. Here I concluded that if x and y are greater than zero (This was not given to me I assumed this), then determinant is greater than zero so <span class="math-container">$U$</span> must be quasi-concave and hence preferences are convex. Is this correct?</p>
<p>For part 2, I considered three cases. <br />
<br />
Case 1: when <span class="math-container">$z = 0$</span> and <span class="math-container">$x,y &gt; 0$</span>. This was just the Walrasian for the standard Cobb Douglas we are left with.<br />
<br />
Case 2: when <span class="math-container">$x$</span> or <span class="math-container">$y = 0$</span> and <span class="math-container">$z &gt; 0$</span>. All wealth is spent on z as well. <br />
<br />
Case 3:  <span class="math-container">$x,y,z &gt; 0$</span>. This I was unable to compute and did not know how to proceed.</p>
<p>Note: Budget constraint is standard <span class="math-container">$P_1X+P_2Y+P_3Z = W$</span></p>
<p>For Case 3: I set up Lagrangian and used Kuhn Tucker Conditions:</p>
<ol>
<li><span class="math-container">$\frac{2}{3}$</span>*<span class="math-container">$(\frac{y}{x})^{\frac{1}{3}} - \lambda P_1 \le 0$</span></li>
<li><span class="math-container">$\frac{1}{3}$</span>*<span class="math-container">$(\frac{x}{y})^{\frac{2}{3}} - \lambda P_2 \le 0$</span></li>
<li><span class="math-container">$1 - \lambda P_3 \le 0$</span></li>
<li><span class="math-container">$x[\frac{2}{3}$</span>*<span class="math-container">$(\frac{y}{x})^{\frac{1}{3}} - \lambda P_1] = 0$</span></li>
<li><span class="math-container">$y[\frac{1}{3}$</span>*<span class="math-container">$(\frac{x}{y})^{\frac{2}{3}} - \lambda P_2] = 0$</span></li>
<li><span class="math-container">$z[1 - \lambda P_3] = 0$</span></li>
<li><span class="math-container">$W - P_1X - P_2Y - P_3Z \ge 0 $</span></li>
<li><span class="math-container">$\lambda [W - P_1X - P_2Y - P_3Z] = 0$</span></li>
</ol>
<p>Imposing <span class="math-container">$x, y, z &gt; 0$</span> and Walras's law, I know that I can equate 1, 2, 3, and 7 to zero. Essentially I end up with an equation which says that at optimal Marginal Utility to Price Ratio of each good is same.</p>
<p>After simplifying by equating <span class="math-container">$\lambda$</span>, I get this from 1, 2:</p>
<p><span class="math-container">$\frac{x}{P_2} = \frac{2y}{P_1}$</span>. <br />
My problem is that I can't solve for <span class="math-container">$z$</span> as I can't get my budget constraint all in terms of one variable.</p>

CC BY-SA 4.0
2021-11-16T18:17:28.457Z
In the part 2 I think you may need to look in cases when the price is different, because the good $z$ is like a substitute of the combination of $x & y$ so if the price of $z$ is much higher than the prices for the combination $x & y$ the consumer will consume only $x & y$ solving the demand as a standard coub douglas, while if the price of $z$ is much lower than the prices for the combination $x & y$ the consumer will consume only $z$.
Macosso
<microeconomics><mathematical-economics><consumer-theory><walrasian>
Solve for the Walrasian demand, Utility of three variables, and Convexity of Preferences?
Kinno
# Solve for the Walrasian demand, Utility of three variables, and Convexity of Preferences?

(Asked by: username_0 on 2021-11-16)

I am given $U(x,y,z) = x^\frac{2}{3}y^\frac{1}{3} + z$. I am asked to solve the following:

(i) Prove the convexity of these preferences (convex, strictly convex or neither?)  
(ii) Solve for the Walrasian Demand?

For part 1, I calculated the determinant of the bordered hessian matrix and got $\frac{320}{81}\*\frac{1}{x^\frac{2}{3}y^\frac{4}{3}}$. Here I concluded that if x and y are greater than zero (This was not given to me I assumed this), then determinant is greater than zero so $U$ must be quasi-concave and hence preferences are convex. Is this correct?

For part 2, I considered three cases.   
  
Case 1: when $z = 0$ and $x,y > 0$. This was just the Walrasian for the standard Cobb Douglas we are left with.  
  
Case 2: when $x$ or $y = 0$ and $z > 0$. All wealth is spent on z as well.   
  
Case 3: $x,y,z > 0$. This I was unable to compute and did not know how to proceed.

Note: Budget constraint is standard $P\_1X+P\_2Y+P\_3Z = W$

For Case 3: I set up Lagrangian and used Kuhn Tucker Conditions:

1. $\frac{2}{3}$\*$(\frac{y}{x})^{\frac{1}{3}} - \lambda P\_1 \le 0$
2. $\frac{1}{3}$\*$(\frac{x}{y})^{\frac{2}{3}} - \lambda P\_2 \le 0$
3. $1 - \lambda P\_3 \le 0$
4. $x[\frac{2}{3}$\*$(\frac{y}{x})^{\frac{1}{3}} - \lambda P\_1] = 0$
5. $y[\frac{1}{3}$\*$(\frac{x}{y})^{\frac{2}{3}} - \lambda P\_2] = 0$
6. $z[1 - \lambda P\_3] = 0$
7. $W - P\_1X - P\_2Y - P\_3Z \ge 0 $
8. $\lambda [W - P\_1X - P\_2Y - P\_3Z] = 0$

Imposing $x, y, z > 0$ and Walras's law, I know that I can equate 1, 2, 3, and 7 to zero. Essentially I end up with an equation which says that at optimal Marginal Utility to Price Ratio of each good is same.

After simplifying by equating $\lambda$, I get this from 1, 2:

$\frac{x}{P\_2} = \frac{2y}{P\_1}$.   
My problem is that I can't solve for $z$ as I can't get my budget constraint all in terms of one variable.

**Tags:** microeconomics, mathematical-economics, consumer-theory, walrasian

> **Comments:**
> **username_1**: In the part 2 I think you may need to look in cases when the price is different, because the good $z$ is like a substitute of the combination of $x & y$ so if the price of $z$ is much higher than the prices for the combination $x & y$ the consumer will consume only $x & y$ solving the demand as a standard coub douglas, while if the price of $z$ is much lower than the prices for the combination $x & y$ the consumer will consume only $z$.  

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## 2 answers

### Answer by username_2 on 2021-11-16. Score: 2 | Accepted answer

For part (i), in complete rigor, you should also check [the determinants of all the leading principal minors of the bordered Hessian and make sure that they have alternating signs](https://mjo.osborne.economics.utoronto.ca/index.php/tutorial/index/1/qcc/t#d:borderedHessian). Your final conclusion looks correct though.

For part (ii), recall that Walrasian demand is the solution to utility maximization subject to budget constraint. So you should setup a Lagrangian, derive the [Kuhn-Tucker conditions](https://mjo.osborne.economics.utoronto.ca/index.php/tutorial/index/1/ktc/t), and then solve for $x,y,z$ as functions of the prices and income. These will be the Walrasian demands.

> **Comments:**
> **username_0**: Thanks for your response. I have updated my question with the Kuhn Tucker conditions. Because z, disappears I am unable to isolate x, y, and z just in terms of prices and wealth, which is what the Walrasian correspondence demands.  
> **username_2**: @username_0: You can usually recover $z$ from the budget constraint after solving for $x$ and $y$. You should probably start from your condition $3$, and use the fact/assumption that $p_1,p_2,p_3>0$ to deduce that $\lambda>0$, which in turn will give you strict equality in condition $7$, and go from there.  
> **username_2**: @username_0: Also, the MU-to-Price Ratio gives you a relation between $x$ and $y$, e.g. $x=2yp_2/p_1$. Plug this into either $1$ or $2$ for $x$, and you'll get $y$ as an expression of prices and $\lambda$.  
> **username_3**: When all are positive x,y,z>0 there is an infinity of solutions. This only happens at particular prices. Walras demand is in this knife-edge case not strictly speaking a function of price and income as suggested by @username_2.  
> **username_2**: @username_3: Yes you're right. Thanks for pointing that out.  

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### Answer by username_4 on 2023-10-15. Score: 1

$u(x,y,z)=x^{\frac{2}{3}}y^{\frac{1}{3}}+z$ is the sum of two concave functions: $u\_1(x,y,z)=x^{\frac{2}{3}}y^{\frac{1}{3}}$ and $u\_2(x,y,z)=z$, and therefore, it is concave. Since $u$ is concave, it is also quasi-concave. Preferences represented by $u$ are, therefore, convex. However, these preferences are not strictly convex. To show this, consider $(x\_1,y\_1,z\_1)=(2,2,0)$ and $(x\_2,y\_2,z\_2)=(0,0,2)$, and $\lambda = \frac{1}{2}$. Clearly,
$u(\lambda (x\_1,y\_1,z\_1) + (1-\lambda)(x\_2,y\_2,z\_2))=u(1,1,1)=2=u(2,2,0)=u(0,0,2)=\min(u(x\_1,y\_1,z\_1),u(x\_2,y\_2,z\_2))$

Demand can be obtained by solving this problem:
\begin{eqnarray\*} \max\_{x,y,z} & \ x^{\frac{2}{3}}y^{\frac{1}{3}}+z \\ \text{s.t. } & p\_Xx+p\_Yy+p\_Zz\leq M \\ \text{and } & x\geq 0, \ y\geq 0, \ z \geq 0\end{eqnarray\*}
where $p\_X>0, p\_Y>0, p\_Z>0, M\geq 0$

and we get the demand as

$(x^d,y^d,z^d)(p\_X,p\_Y,p\_Z,M)\in\begin{cases} \left\{\left(\dfrac{2M}{3p\_X},\dfrac{M}{3p\_Y},0\right)\right\} &\text{if } \frac{3p\_X^{\frac{2}{3}}p\_Y^{\frac{1}{3}}}{2^{\frac{2}{3}}}<p\_Z \\ \left\{\left(0,0,\dfrac{M}{p\_Z}\right)\right\} & \text{if } \frac{3p\_X^{\frac{2}{3}}p\_Y^{\frac{1}{3}}}{2^{\frac{2}{3}}}>p\_Z \\ \left\{\left(\dfrac{2\alpha M}{3p\_X},\dfrac{\alpha M}{3p\_Y},\dfrac{(1-\alpha)M}{p\_Z}\right)|0\leq\alpha\leq 1\right\}&\text{if } \frac{3p\_X^{\frac{2}{3}}p\_Y^{\frac{1}{3}}}{2^{\frac{2}{3}}}=p\_Z\end{cases}$

Note: Approach I have used to find the demand here is same as the one I have used here: <https://economics.stackexchange.com/a/51562/11824>


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