I would like to propose to you the following problem and my proposed solution. In particular, I am unsure in how to correctly characterize the Walrasian demand. Can you please have a look at it and express your opinions and corrections?

Let ${x}=(x_1,x_2)$ denote the consumption vector, ${p}=(p_1,p_2)$ denote the price vector, and let $w$ be the consumer's wealth. The utility maximization problem is

\begin{equation*} \max_{{x\geq 0}} \ \ (x_1 + 1)^\alpha (x_2+1)^\beta \ \ \ \text{ s.t. } {p \cdot x} \leq w.\end{equation*}

with $\alpha + \beta = 1$ with $\alpha, \beta > 0$. The Lagrangian function for the UMP is

\begin{equation*} \mathcal{L} ({x}, \lambda, {\mu} ) = (x_1 + 1)^\alpha (x_2+1)^\beta- \lambda ({p \cdot x} -w) + {\mu \cdot x},\end{equation*}

where $\lambda$ and ${\mu} = (\mu_1,\mu_2)$ are the Lagrangian multipliers.

The Kuhn-Tucker necessary conditions for the consumer problem are formed by the following conditions:

\begin{equation*} \begin{aligned} \frac{\partial \mathcal{L}}{\partial x_1} = \alpha(x_1+1)^{\alpha-1} (x_2 +1)^\beta - \lambda p_1 + \mu_1&= 0,\\[5pt] \frac{\partial \mathcal{L}}{\partial x_2} = \beta(x_2+1)^{\beta-1} (x_1 +1)^\alpha - \lambda p_2 + \mu_2 &= 0, \end{aligned} \end{equation*} with $\lambda \geq 0$ and ${\mu} \geq 0$; the initial constraints,

\begin{equation} {p \cdot x} \leq w \ \ \ \text{ and } \ \ \ {x } \gg 0,\end{equation}

and the complementary slackness conditions,

\begin{equation*} \lambda \frac{\partial \mathcal{L}}{\partial \lambda} = \lambda ({p \cdot x} - w)=0 \ \ \ \ \text{ and } \ \ \ \ \mu_i \frac{\partial \mathcal{L}}{\partial \mu_i} = \mu_i x_i = 0 \ \text{ for } i=1,2. \end{equation*}

Note that under these conditions $\partial u({x})/\partial x_i \leq \lambda p_i$. Since $\nabla u({x}) \gg 0$ and ${p} \gg 0$, this implies $\lambda \geq [\partial u({x})/\partial x_i]/p_i > 0$. In words, the Lagrangian multiplier $\lambda$ is positive and the constraint is binding. Therefore, by $\lambda ({p \cdot x} - w)=0$, we can conclude that ${p \cdot x} = w$ (Walras' law holds).

Interior solutions are of the type ${x} \gg 0$ which imply ${\mu} = {0}$. As we have shown above, because $u({x})$ is an increasing function and since we assume ${p} \gg 0$, Walras' law holds. Thus, the Kuhn-Tucker conditions restrict to the system of equations

$$\begin{align} &\alpha(x_1+1)^{\alpha-1} (x_2 +1)^\beta = \lambda p_1, \qquad &(1)\\[5pt] &\beta(x_2+1)^{\beta-1} (x_1 +1)^\alpha = \lambda p_2, \qquad &(2)\\[5pt] &{x} \gg 0, \qquad &(3)\\[5pt] &{p \cdot x} = w. \qquad &(4) \end{align}$$

By dividing (1) by (2) we obtain the key (tangency) condition for an optimum: the marginal rate of substitution of good 1 with good 2 at the optimum must equal the price ratio of the two goods

\begin{equation*}  \frac{\alpha(x_1+1)^{\alpha-1} (x_2 +1)^\beta}{\beta(x_2+1)^{\beta-1} (x_1 +1)^\alpha}= \frac{p_1}{p_2} \end{equation*}

\begin{equation} \frac{\alpha (x_2 +1)}{\beta(x_1+1)}= \frac{p_1}{p_2} .\qquad (5)\end{equation}

Solving (5) for $x_2$ allows to rewrite the necessary (and sufficient) condition for ${x}=(x_1,x_2) $ to be an optimum in a useful way for our computations

\begin{equation} x_2= \frac{\beta p_1}{\alpha p_2}(x_1+1) -1. \qquad (6) \end{equation}

By substituting (6) into (4) we are now able to solve for $x_1 ({p}, w)$

$$\begin{align} &p_1 x_1 +p_2 \bigg(\frac{\beta p_1}{\alpha p_2}(x_1+1) -1\bigg) = w \nonumber \\[5pt] &p_1 x_1 = \bigg(w - \frac{\beta }{\alpha}p_1 + p_2 \bigg)\frac{\alpha}{\alpha + \beta} \nonumber \\[5pt] &x_1 ({p}, w) = \frac{\alpha (w+p_2)}{p_1} - \beta , \qquad \qquad (7) \end{align}$$

and by substituting (7) into (6) we cans solve for $x_2 ({p}, w)$ \begin{equation*} x_2 ({p}, w) = \frac{\beta (w+p_1)}{p_2} - \alpha .\end{equation*}

For boundary solutions we need to look at the cases where $x_1 = 0$ or $x_2 = 0$ (having both equal zero is uninteresting and certainly not the case for any locally nonsatiated function). The complementary slackness condition $\mu_i x_i = 0 $ for $i=1,2$ implies in the first case $\mu_1 \geq 0$ and in the second $\mu_2 \geq 0$. As before, because $u({x})$ is an increasing function and since we assume ${p} \gg 0$, Walras' law holds (in both cases). \[-7pt]

In the first case, $x_1 = 0$ implies $\mu_1 \geq 0$. Thus, the Kuhn-Tucker conditions restrict to the system of equations

$$\begin{align} &\alpha(x_1+1)^{\alpha-1} (x_2 +1)^\beta \leq \lambda p_1, \qquad &(8)\\[5pt] &\beta(x_2+1)^{\beta-1} (x_1 +1)^\alpha = \lambda p_2,\qquad &(9)\\[5pt] &x_1 = 0, \ \ x_2 > 0, \qquad &(10)\\[5pt] &{p \cdot x} = w. \qquad &(11) \end{align}$$

By substituting the equation in (10) into the budget constraint (11) we obtain the candidate for an optimum, $x_2=w/p_2$. Then, if we divide (8) by (9), and consider $x_1 = 0$, we obtain the necessary (and sufficient) condition for $x_2=w/p_2$ being optimal\footnote{Note that, for boundary solutions, the indifference curve need not be tangent to the budget line.}

\begin{equation*} \frac{\alpha (x_2 +1)}{\beta(x_1+1)} \leq \frac{p_1}{p_2} \end{equation*}

\begin{equation} \frac{\alpha (x_2 +1)}{\beta} \leq \frac{p_1}{p_2}. \end{equation}

Since $\alpha (x_2 +1) / \beta > 0$, there exists some price vector ${p \gg 0}$ that satisfies (12). Hence $x_2=w/p_2$ is optimal for those price vectors ${p}$ that satisfy (12).

In the second case, $x_2 = 0$ implies $\mu_2 \geq 0$. This time we omit most of the algebra and explanation because they bear close similarity to the case above. Jumping to the conclusion: by the Walras' law the candidate for the optimum is $x_1=w/p_1$ and the necessary condition for it being an optimum is

\begin{equation} \frac{\alpha}{\beta(x_1 +1)} \geq \frac{p_1}{p_2}. \end{equation}

Since $\alpha / (x_1 +1)\beta > 0$, there exists some price vector ${p \gg 0}$ that satisfies (12). Hence $x_1=w/p_1$ is optimal for for price vectors ${p}$ that satisfy (13).

Finally, the Walrasian demand in compact notation is

\begin{equation} x({p}, w) = \begin{cases} \bigg(\dfrac{w}{p_1}, 0\bigg) \quad &\text{ if } \ \dfrac{p_1}{p_2} \leq  \dfrac{\alpha}{\beta(x_1 +1)} \\[5pt] \bigg(\dfrac{\alpha (w+p_2)}{p_1} - \beta, \, \dfrac{\beta (w+p_1)}{p_2} - \alpha \bigg) \quad  &\text{ if } \ \dfrac{\alpha}{\beta(x_1 +1)} < \dfrac{p_1}{p_2} < \dfrac{\alpha (x_2 +1)}{\beta}\\[5pt] \bigg(0,\dfrac{w}{p_2}\bigg) \quad &\text{ if } \ \dfrac{p_1}{p_2} \geq \dfrac{\alpha (x_2 +1)}{\beta} \end{cases} \end{equation}

  • $\begingroup$ What are your assumptions on $\alpha$ and $\beta$? $\endgroup$ Commented Dec 14, 2020 at 10:25
  • $\begingroup$ Sorry, I forgot to mention them. $\alpha + \beta = 1$ with $\alpha, \beta > 0$. $\endgroup$
    – fennel
    Commented Dec 14, 2020 at 10:31
  • $\begingroup$ Ok, I think your solution is nice and very well described. Personally I would have solved it using that it is a translation of Cobb-Douglas preferences. I believe that should be possible. $\endgroup$ Commented Dec 14, 2020 at 10:34
  • $\begingroup$ Thanks for your answer. About the Walrasian demand, I could have expressed the conditions on the right as a function of wealth rather than as a function of $x$. My thought is: the Walrasian demand should depend on $p$ and $w$ and having some $x$ in the conditions is in someway "recursive". What do you think? $\endgroup$
    – fennel
    Commented Dec 14, 2020 at 10:41
  • $\begingroup$ I would agree, that would be nicer. I posted the translation strategy as answer because it was too long for comment. $\endgroup$ Commented Dec 14, 2020 at 10:44

1 Answer 1


To solve $$\max_{x \geq 0} \ (x_1+1)^\alpha(x_2 + 1)^\beta$$

$$s.t. \ \ I \geq p_1x_1 + p_2x_2,$$

I would define $y_1 = x_1+1$ and $y_2 = x_2 + 1$ to get the problem

$$\max_{y \geq 0} \ y_1^\alpha y_2^\beta$$

$$s.t. \ \ \bar I \geq p_1y_1 + p_2y_2,$$

where $\bar I := I + p_1 + p_2$. For $\alpha + \beta = 1$ the solution is well known to be

$$y_1^* = \frac{\alpha \bar I}{p_1},$$

which given the definitions imply that

$$x^*_1 + 1 = \frac{\alpha(I + p_1 + p_2)}{p_1} \Leftrightarrow \\[8pt] x^*_1 = \frac{\alpha(I + p_2)}{p_1} - \beta$$

My guess would be the condition for a border solution is that

$$x_1^* \leq 0 \Leftrightarrow \frac{\alpha(I + p_2)}{p_1} - \beta \leq 0$$

which is equivalent to

$$I \leq \frac{\beta p_1}{\alpha} - p_2.$$

Does this make any sense? Well it does in so far that if $p_1$ is very high then consumer want buy anything of $x_1$.

  • $\begingroup$ This trick is nice, but what about eventual boundary solutions? $\endgroup$
    – fennel
    Commented Dec 14, 2020 at 10:45
  • $\begingroup$ Well again depending on knowledge of CD preferences you know indefference curves never cut 0 but moving the origin to (1,1) indifference curves can cut axis, but that will simply be as you find: All income to one good solution. I have not however given any thought to how to find the conditions for when that would occur. So I agree as it stands this strategy is incomplete. $\endgroup$ Commented Dec 14, 2020 at 10:48
  • 1
    $\begingroup$ I guess you can derive the condition for $x_2$ as border solution by setting the solution found for $x_1 \leq 0$ $\endgroup$ Commented Dec 14, 2020 at 10:58

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