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.

  • $\begingroup$ 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 $z$. $\endgroup$
    – Macosso
    Nov 16, 2021 at 18:17

2 Answers 2


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. 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, and then solve for $x,y,z$ as functions of the prices and income. These will be the Walrasian demands.

  • $\begingroup$ 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. $\endgroup$
    – Kinno
    Nov 17, 2021 at 11:27
  • $\begingroup$ @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. $\endgroup$
    – Herr K.
    Nov 17, 2021 at 13:39
  • $\begingroup$ @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 $\lambda$. $\endgroup$
    – Herr K.
    Nov 17, 2021 at 13:48
  • 1
    $\begingroup$ 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. $\endgroup$ Nov 24, 2021 at 23:07
  • $\begingroup$ @JesperHybel: Yes you're right. Thanks for pointing that out. $\endgroup$
    – Herr K.
    Nov 25, 2021 at 1:55

$u(x,y,z)=x^{\frac{2}{3}}y^{\frac{1}{3}}+z$ 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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