<p>To solve
<span class="math-container">$$\max_{x \geq 0} \ (x_1+1)^\alpha(x_2 + 1)^\beta$$</span></p>
<p><span class="math-container">$$s.t. \ \ I \geq p_1x_1 + p_2x_2,$$</span></p>
<p>I would define <span class="math-container">$y_1 = x_1+1$</span> and <span class="math-container">$y_2 = x_2 + 1$</span> to get the problem</p>
<p><span class="math-container">$$\max_{y \geq 0} \ y_1^\alpha y_2^\beta$$</span></p>
<p><span class="math-container">$$s.t. \ \ \bar I \geq p_1y_1 + p_2y_2,$$</span></p>
<p>where <span class="math-container">$\bar I := I + p_1 + p_2$</span>. For <span class="math-container">$\alpha + \beta = 1$</span> the solution is well known to be</p>
<p><span class="math-container">$$y_1^* = \frac{\alpha \bar I}{p_1},$$</span></p>
<p>which given the definitions imply that</p>
<p><span class="math-container">$$x^*_1 + 1 = \frac{\alpha(I + p_1 + p_2)}{p_1} \Leftrightarrow \\[8pt]
x^*_1 = \frac{\alpha(I + p_2)}{p_1} - \beta$$</span></p>
<p>My guess would be the condition for a border solution is that</p>
<p><span class="math-container">$$x_1^* \leq 0 \Leftrightarrow  \frac{\alpha(I + p_2)}{p_1} - \beta \leq 0$$</span></p>
<p>which is equivalent to</p>
<p><span class="math-container">$$I  \leq \frac{\beta p_1}{\alpha} - p_2.$$</span></p>
<p>Does this make any sense? Well it does in so far that if <span class="math-container">$p_1$</span> is very high then consumer want buy anything of <span class="math-container">$x_1$</span>.</p>

CC BY-SA 4.0
2020-12-14T10:45:27.327Z
This trick is nice, but what about eventual boundary solutions?
fennel
CC BY-SA 4.0
2020-12-14T10:48:43.047Z
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.
Jesper Hybel
CC BY-SA 4.0
2020-12-14T10:58:13.393Z
I guess you can derive the condition for $x_2$ as border solution by setting the solution found for $x_1 \leq 0$
Jesper Hybel
CC BY-SA 4.0
2020-12-14T10:43:01.437Z
2020-12-14T11:05:41.603Z
2020-12-14T11:05:41.603Z
Jesper Hybel
<p>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?</p>
<p>Let <span class="math-container">${x}=(x_1,x_2)$</span> denote the consumption vector, <span class="math-container">${p}=(p_1,p_2)$</span> denote the price vector, and let <span class="math-container">$w$</span> be the consumer's wealth. The utility maximization problem is</p>
<p><span class="math-container">\begin{equation*} \max_{{x\geq 0}} \ \  (x_1 + 1)^\alpha (x_2+1)^\beta \ \ \ \text{ s.t. } {p \cdot x} \leq w.\end{equation*}</span></p>
<p>with <span class="math-container">$\alpha + \beta = 1$</span> with <span class="math-container">$\alpha, \beta &gt; 0$</span>. The Lagrangian function for the UMP is</p>
<p><span class="math-container">\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*}</span></p>
<p>where <span class="math-container">$\lambda$</span> and <span class="math-container">${\mu} = (\mu_1,\mu_2)$</span> are the Lagrangian multipliers.</p>
<p>The Kuhn-Tucker necessary conditions for the consumer problem are formed by the following conditions:</p>
<p><span class="math-container">\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&amp;= 0,\\[5pt]
        \frac{\partial  \mathcal{L}}{\partial x_2} = \beta(x_2+1)^{\beta-1} (x_1 +1)^\alpha - \lambda p_2 + \mu_2 &amp;= 0,
    \end{aligned}
\end{equation*}</span>
with <span class="math-container">$\lambda \geq 0$</span> and <span class="math-container">${\mu} \geq 0$</span>; the initial constraints,</p>
<p><span class="math-container">\begin{equation} {p \cdot x} \leq w \ \ \ \text{ and } \ \ \ {x } \gg 0,\end{equation}</span></p>
<p>and the complementary slackness conditions,</p>
<p><span class="math-container">\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*}</span></p>
<p>Note that under these conditions <span class="math-container">$\partial u({x})/\partial x_i  \leq \lambda p_i$</span>. Since <span class="math-container">$\nabla u({x}) \gg 0$</span> and <span class="math-container">${p} \gg 0$</span>, this implies <span class="math-container">$\lambda \geq [\partial u({x})/\partial x_i]/p_i &gt; 0$</span>. In words, the Lagrangian multiplier <span class="math-container">$\lambda$</span> is positive and the constraint is binding. Therefore, by  <span class="math-container">$\lambda ({p \cdot x} - w)=0$</span>, we can conclude that  <span class="math-container">${p \cdot x} = w$</span> (Walras' law holds).</p>
<p>Interior solutions are of the type <span class="math-container">${x} \gg 0$</span> which imply
<span class="math-container">${\mu} = {0}$</span>. As we have shown above, because <span class="math-container">$u({x})$</span> is an increasing function and since we assume <span class="math-container">${p} \gg 0$</span>, Walras' law holds. Thus, the Kuhn-Tucker conditions restrict to the system of equations</p>
<p><span class="math-container">$$\begin{align}
        &amp;\alpha(x_1+1)^{\alpha-1} (x_2 +1)^\beta  = \lambda p_1, \qquad &amp;(1)\\[5pt]
        &amp;\beta(x_2+1)^{\beta-1} (x_1 +1)^\alpha = \lambda p_2, \qquad &amp;(2)\\[5pt]
        &amp;{x} \gg 0, \qquad &amp;(3)\\[5pt]
        &amp;{p \cdot x} = w. \qquad &amp;(4)
\end{align}$$</span></p>
<p>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</p>
<p><span class="math-container">\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*}</span></p>
<p><span class="math-container">\begin{equation} \frac{\alpha (x_2 +1)}{\beta(x_1+1)}= \frac{p_1}{p_2} .\qquad (5)\end{equation}</span></p>
<p>Solving (5) for <span class="math-container">$x_2$</span> allows to rewrite the necessary (and sufficient) condition for <span class="math-container">${x}=(x_1,x_2) $</span> to be an optimum in a useful way for our computations</p>
<p><span class="math-container">\begin{equation}
    x_2= \frac{\beta p_1}{\alpha p_2}(x_1+1) -1. \qquad (6)
\end{equation}</span></p>
<p>By substituting (6) into (4) we are now able to solve for <span class="math-container">$x_1 ({p}, w)$</span></p>
<p><span class="math-container">$$\begin{align}
        &amp;p_1 x_1 +p_2 \bigg(\frac{\beta p_1}{\alpha p_2}(x_1+1) -1\bigg) = w \nonumber \\[5pt] 
        &amp;p_1 x_1 = \bigg(w - \frac{\beta }{\alpha}p_1 + p_2 \bigg)\frac{\alpha}{\alpha + \beta} \nonumber \\[5pt]
        &amp;x_1 ({p}, w) = 
         \frac{\alpha (w+p_2)}{p_1} - \beta , \qquad \qquad (7)
\end{align}$$</span></p>
<p>and by substituting (7) into (6) we cans solve for <span class="math-container">$x_2 ({p}, w)$</span>
<span class="math-container">\begin{equation*} x_2 ({p}, w) = \frac{\beta (w+p_1)}{p_2} - \alpha .\end{equation*}</span></p>
<p>For boundary solutions we need to look at the cases where <span class="math-container">$x_1 = 0$</span> or <span class="math-container">$x_2 = 0$</span> (having both equal zero is uninteresting and certainly not the case for any locally nonsatiated function). The complementary slackness condition <span class="math-container">$\mu_i x_i = 0 $</span> for <span class="math-container">$i=1,2$</span> implies in the first case <span class="math-container">$\mu_1 \geq 0$</span> and in the second <span class="math-container">$\mu_2 \geq 0$</span>. As before, because <span class="math-container">$u({x})$</span> is an increasing function and since we assume <span class="math-container">${p} \gg 0$</span>, Walras' law holds (in both cases). \[-7pt]</p>
<p>In the first case, <span class="math-container">$x_1 = 0$</span> implies <span class="math-container">$\mu_1 \geq 0$</span>.  Thus, the Kuhn-Tucker conditions restrict to the system of equations</p>
<p><span class="math-container">$$\begin{align}
    &amp;\alpha(x_1+1)^{\alpha-1} (x_2 +1)^\beta  \leq \lambda p_1, \qquad &amp;(8)\\[5pt]
    &amp;\beta(x_2+1)^{\beta-1} (x_1 +1)^\alpha = \lambda p_2,\qquad &amp;(9)\\[5pt]
    &amp;x_1 = 0, \ \ x_2 &gt; 0, \qquad &amp;(10)\\[5pt]
    &amp;{p \cdot x} = w. \qquad &amp;(11)
\end{align}$$</span></p>
<p>By substituting the equation in (10) into the budget constraint (11) we obtain the candidate for an optimum, <span class="math-container">$x_2=w/p_2$</span>.
Then, if we divide (8) by (9), and consider <span class="math-container">$x_1 = 0$</span>, we obtain the necessary (and sufficient) condition for <span class="math-container">$x_2=w/p_2$</span> being optimal\footnote{Note that, for boundary solutions, the indifference curve need not be tangent to the budget line.}</p>
<p><span class="math-container">\begin{equation*} \frac{\alpha (x_2 +1)}{\beta(x_1+1)} \leq \frac{p_1}{p_2} \end{equation*}</span></p>
<p><span class="math-container">\begin{equation}
\frac{\alpha (x_2 +1)}{\beta} \leq \frac{p_1}{p_2}.
\end{equation}</span></p>
<p>Since <span class="math-container">$\alpha (x_2 +1) / \beta &gt; 0$</span>, there exists some price vector <span class="math-container">${p \gg 0}$</span> that satisfies (12). Hence <span class="math-container">$x_2=w/p_2$</span> is optimal for those price vectors <span class="math-container">${p}$</span> that satisfy (12).</p>
<p>In the second case, <span class="math-container">$x_2 = 0$</span> implies <span class="math-container">$\mu_2 \geq 0$</span>. 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 <span class="math-container">$x_1=w/p_1$</span> and the necessary condition for it being an optimum is</p>
<p><span class="math-container">\begin{equation} \frac{\alpha}{\beta(x_1 +1)} \geq \frac{p_1}{p_2}. \end{equation}</span></p>
<p>Since <span class="math-container">$\alpha /  (x_1 +1)\beta &gt; 0$</span>, there exists some price vector <span class="math-container">${p \gg 0}$</span> that satisfies (12). Hence <span class="math-container">$x_1=w/p_1$</span> is optimal for for price vectors <span class="math-container">${p}$</span> that satisfy (13).</p>
<p>Finally, the Walrasian demand in compact notation is</p>
<p><span class="math-container">\begin{equation}
x({p}, w) = 
\begin{cases}
    \bigg(\dfrac{w}{p_1}, 0\bigg) \quad &amp;\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  &amp;\text{ if } \ \dfrac{\alpha}{\beta(x_1 +1)} &lt; \dfrac{p_1}{p_2} &lt;  \dfrac{\alpha (x_2 +1)}{\beta}\\[5pt]
    \bigg(0,\dfrac{w}{p_2}\bigg) \quad &amp;\text{ if } \ \dfrac{p_1}{p_2}  \geq \dfrac{\alpha (x_2 +1)}{\beta}
\end{cases}
\end{equation}</span></p>

CC BY-SA 4.0
2020-12-14T10:25:52.580Z
What are your assumptions on $\alpha$ and $\beta$?
Jesper Hybel
CC BY-SA 4.0
2020-12-14T10:31:45.957Z
Sorry, I forgot to mention them. $\alpha + \beta = 1$ with $\alpha, \beta > 0$.
fennel
CC BY-SA 4.0
2020-12-14T10:34:29.433Z
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.
Jesper Hybel
CC BY-SA 4.0
2020-12-14T10:41:46.550Z
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?
fennel
CC BY-SA 4.0
2020-12-14T10:44:33.213Z
I would agree, that would be nicer. I posted the translation strategy as answer because it was too long for comment.
Jesper Hybel
<microeconomics><consumer-theory><demand><walrasian>
Correct and complete characterisation of the Walrasian demand function
fennel
# Correct and complete characterisation of the Walrasian demand function

(Asked by: username_0 on 2020-12-14)

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}

**Tags:** microeconomics, consumer-theory, demand, walrasian
> **Comments:**
> **username_1**: What are your assumptions on $\alpha$ and $\beta$?  
> **username_0**: Sorry, I forgot to mention them. $\alpha + \beta = 1$ with $\alpha, \beta > 0$.  
> **username_1**: 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.  
> **username_0**: 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?  
> **username_1**: I would agree, that would be nicer. I posted the translation strategy as answer because it was too long for comment.  

---

## 1 answers

### Answer by username_1 on 2020-12-14. Score: 5 | Accepted answer

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 $x\_1$.

> **Comments:**
> **username_0**: This trick is nice, but what about eventual boundary solutions?  
> **username_1**: 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.  
> **username_1**: I guess you can derive the condition for $x_2$ as border solution by setting the solution found for $x_1 \leq 0$  

---

