<p>An interesting issue arose while exchanging comments related to @MichaelGreineckeranswer. So let's explore this formally</p>
<blockquote>
<p>Claim: There is no boundary <em>solution</em> with
<span class="math-container">$x_1 = 0$</span>.</p>
</blockquote>
<p>Budget exhaustion holds, so with <span class="math-container">$x_1 = 0$</span> we will have <span class="math-container">$U=x_2 = w/p_2$</span>. So in an attempt to disprove the claim, we examine under which conditions the following inequality will hold:</p>
<p><span class="math-container">$$\frac{w}{p_2} &gt; x_1^{\alpha} + x_2,\quad p_1x_1 + p_2x_2 = w.$$</span></p>
<p>Using the budget constraint to solve for <span class="math-container">$x_2$</span> we can rewrite this</p>
<p><span class="math-container">$$\frac{w}{p_2} &gt; x_1^{\alpha} + \frac{w-p_1x_1}{p_2}\;\implies\; \frac{p_1}{p_2}x_1 &gt; x_1^{\alpha} \implies x_1^{1-\alpha} &gt; \frac{p_2}{p_1}$$</span></p>
<p><span class="math-container">$$ \implies  x_1 &gt; \left(\frac{p_2}{p_1}\right)^{1/(1-\alpha)} \tag{1}$$</span></p>
<p>So we have obtained: The utility from consuming <span class="math-container">$x_2$</span> only, will be greater than the utility from any consumption plan for which the consumption bundle includes a quantity of <span class="math-container">$x_1$</span> <em>higher</em> than the right-hand-side of <span class="math-container">$(1)$</span>.</p>
<p>The crucial word her is &quot;higher&quot; -because it implies that bundles that include strictly positive amounts of both <span class="math-container">$(x_1, x_2)$</span> and as long as the amount of <span class="math-container">$x_1$</span> is <em>lower</em> than the right-hand-side of <span class="math-container">$(1)$</span>, are superior in terms of utility than consuming <span class="math-container">$x_2$</span> only.</p>
<p>So @MichaelGreinecker is right: the solution set does not include a bundle with zero <span class="math-container">$x_1$</span>.</p>

CC BY-SA 4.0
2020-12-14T03:11:22.233Z
2020-12-14T03:11:22.233Z
Alecos Papadopoulos
<p>Except for the extreme cases <span class="math-container">$\alpha=0$</span> and <span class="math-container">$\alpha=1$</span>, there is no boundary solution with <span class="math-container">$x_1=0$</span>. Note that the marginal utility of good 2 is constant, while the limiting marginal utility for <span class="math-container">$x_1\to 0$</span> is infinite. So the consumer will always consume a bit of <span class="math-container">$x_1$</span>.</p>

CC BY-SA 4.0
2020-12-13T21:58:35.190Z
@MichaelGreinder Are you sure about that? Utility is additive here, so intuitively, if $x_2$ is sufficiently cheap, the consumer may achieve a higher utility index by spending all their budget in $x_2$.
Alecos Papadopoulos
CC BY-SA 4.0
2020-12-13T22:07:51.233Z
How should I formalize (mathematically) the nonexistence of boundary solutions with $x_1=0$ (when not in extreme cases)?
fennel
CC BY-SA 4.0
2020-12-13T22:47:11.693Z
@Alecos Fix $p_1$ and $p_2$. The marginal utility per unit of money spent on good $2$ is $1/p_2$ at each bundle. Let $\beta=(1-\alpha)>0$. Then the marginal utility from $x_1$ is $\alpha 1/x^\beta$. For $x_1$ small enough, $\alpha 1/(x^\beta p_1)$ is larger than $1/p_2$.
Michael Greinecker
CC BY-SA 4.0
2020-12-13T18:35:23.323Z
2020-12-13T18:35:23.323Z
Michael Greinecker
<p>I'm trying to find boundary solutions for the following utility maximization problem, but i'm unsure on how to proceed. Here is the problem and what I got so far:</p>
<p><span class="math-container">$ \max x_1^\alpha + x_2 \qquad \text{s.t.}\ x_1 \geq 0,\ x_2 \geq 0,\ p_1x_1 + p_2x_2 \leq w$</span></p>
<p>where <span class="math-container">$\alpha \in [0,1]$</span>. The Kuhn-Tucker conditions are</p>
<p><span class="math-container">$\begin{equation}
\dfrac{\partial \mathcal{L}}{\partial x_1} = \alpha x_1 ^{\alpha -1} - \lambda p_1 + \mu_1=0,\\
\dfrac{\partial \mathcal{L}}{\partial x_2} = 1 - \lambda p_2 + \mu_2 =0,\\
p_1x_1 + p_2x_2 \leq w,\\
x_1\geq 0,\\
x_2 \geq 0,\\
\lambda(p_1x_1 + p_2x_2-w)=0,\\
\mu_1x_1 = 0,\\
\mu_2x_2=0,\\
\lambda \geq 0, \quad \mu_1 \geq 0, \quad \mu_2 \geq 0.\\
\end{equation}$</span></p>
<p>If I consider boundary solutions with <span class="math-container">$x_1=0$</span> and <span class="math-container">$x_2&gt;0$</span>, then by the complementary slackness condition we have <span class="math-container">$\mu_1 \geq 0$</span> and <span class="math-container">$\mu_2=0$</span>.</p>
<p>Since <span class="math-container">$\nabla u(x_1,x_2) \gg 0$</span> and <span class="math-container">$p_1,p_2 &gt; 0$</span>, <span class="math-container">$\lambda &gt; 0$</span>. The constraint on the budget set is binding (Walras' law holds). My candidate for the optimum therefore is <span class="math-container">$x_2=w/p_2$</span>.</p>
<p>The Kuhn-Tucker necessary (and in this case also sufficient) conditions for <span class="math-container">$x_2=w/p_2$</span> being optimal restrict to</p>
<p><span class="math-container">$\begin{equation}
 \alpha x_1 ^{\alpha -1} \leq \lambda p_1 ,\\
 1 = \lambda p_2,\\
x_1 = 0,\\
x_2 \geq 0.
\end{equation}$</span></p>
<p>By dividing the first condition by the second we obtain</p>
<p><span class="math-container">$MRS_{1,2}(x_1,x_2)= \alpha x_1 ^{\alpha -1} \leq \dfrac{p_1}{p_2}$</span></p>
<p>but then I do not know how to proceed.</p>

CC BY-SA 4.0
2020-12-13T18:33:51.887Z
Why your optimal candidate includes $p_1$ and not $p_2$?
Alecos Papadopoulos
CC BY-SA 4.0
2020-12-13T21:42:14.317Z
Write first condition not as inequality but as equality including $\mu_1$ then use second equality to remove lambda and let now x1 go to zero ..... or dont do any of that but simply use that lambda is finiteb positive and let x1 go to 0 then first condition becomes absurd and you know that is not solution.
Jesper Hybel
CC BY-SA 4.0
2020-12-13T21:45:05.540Z
@AlecosPapadopoulos it is a typo. Thanks for pointing it out.
fennel
CC BY-SA 4.0
2020-12-13T21:49:43.993Z
@JesperHybel So, the fact that there is no price vector that satisfies the last inequality when $x_1=0$ tells me that no boundary solution is possible?
fennel
CC BY-SA 4.0
2020-12-13T21:53:37.503Z
It tells you that x1=0 x2>0 is not possible. Just like Greinecker says. But you still have chek x2=0 x1>0.
Jesper Hybel
CC BY-SA 4.0
2020-12-13T21:55:40.613Z
Thanks! Yes, the other boundary solution is less problematic than this one.
fennel
<microeconomics><consumer-theory>
Boundary solutions in the Utility Maximization Problem
fennel
# Boundary solutions in the Utility Maximization Problem

(Asked by: username_0 on 2020-12-13)

I'm trying to find boundary solutions for the following utility maximization problem, but i'm unsure on how to proceed. Here is the problem and what I got so far:

$ \max x\_1^\alpha + x\_2 \qquad \text{s.t.}\ x\_1 \geq 0,\ x\_2 \geq 0,\ p\_1x\_1 + p\_2x\_2 \leq w$

where $\alpha \in [0,1]$. The Kuhn-Tucker conditions are

$\begin{equation}
\dfrac{\partial \mathcal{L}}{\partial x\_1} = \alpha x\_1 ^{\alpha -1} - \lambda p\_1 + \mu\_1=0,\\
\dfrac{\partial \mathcal{L}}{\partial x\_2} = 1 - \lambda p\_2 + \mu\_2 =0,\\
p\_1x\_1 + p\_2x\_2 \leq w,\\
x\_1\geq 0,\\
x\_2 \geq 0,\\
\lambda(p\_1x\_1 + p\_2x\_2-w)=0,\\
\mu\_1x\_1 = 0,\\
\mu\_2x\_2=0,\\
\lambda \geq 0, \quad \mu\_1 \geq 0, \quad \mu\_2 \geq 0.\\
\end{equation}$

If I consider boundary solutions with $x\_1=0$ and $x\_2>0$, then by the complementary slackness condition we have $\mu\_1 \geq 0$ and $\mu\_2=0$.

Since $\nabla u(x\_1,x\_2) \gg 0$ and $p\_1,p\_2 > 0$, $\lambda > 0$. The constraint on the budget set is binding (Walras' law holds). My candidate for the optimum therefore is $x\_2=w/p\_2$.

The Kuhn-Tucker necessary (and in this case also sufficient) conditions for $x\_2=w/p\_2$ being optimal restrict to

$\begin{equation}
\alpha x\_1 ^{\alpha -1} \leq \lambda p\_1 ,\\
1 = \lambda p\_2,\\
x\_1 = 0,\\
x\_2 \geq 0.
\end{equation}$

By dividing the first condition by the second we obtain

$MRS\_{1,2}(x\_1,x\_2)= \alpha x\_1 ^{\alpha -1} \leq \dfrac{p\_1}{p\_2}$

but then I do not know how to proceed.

**Tags:** microeconomics, consumer-theory
> **Comments:**
> **username_1**: Why your optimal candidate includes $p_1$ and not $p_2$?  
> **username_2**: Write first condition not as inequality but as equality including $\mu_1$ then use second equality to remove lambda and let now x1 go to zero ..... or dont do any of that but simply use that lambda is finiteb positive and let x1 go to 0 then first condition becomes absurd and you know that is not solution.  
> **username_0**: @AlecosPapadopoulos it is a typo. Thanks for pointing it out.  
> **username_0**: @JesperHybel So, the fact that there is no price vector that satisfies the last inequality when $x_1=0$ tells me that no boundary solution is possible?  
> **username_2**: It tells you that x1=0 x2>0 is not possible. Just like Greinecker says. But you still have chek x2=0 x1>0.  
> **username_0**: Thanks! Yes, the other boundary solution is less problematic than this one.  

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## 2 answers

### Answer by username_3 on 2020-12-13. Score: 5 | Accepted answer

Except for the extreme cases $\alpha=0$ and $\alpha=1$, there is no boundary solution with $x\_1=0$. Note that the marginal utility of good 2 is constant, while the limiting marginal utility for $x\_1\to 0$ is infinite. So the consumer will always consume a bit of $x\_1$.

> **Comments:**
> **username_1**: @MichaelGreinder Are you sure about that? Utility is additive here, so intuitively, if $x_2$ is sufficiently cheap, the consumer may achieve a higher utility index by spending all their budget in $x_2$.  
> **username_0**: How should I formalize (mathematically) the nonexistence of boundary solutions with $x_1=0$ (when not in extreme cases)?  
> **username_3**: @Alecos Fix $p_1$ and $p_2$. The marginal utility per unit of money spent on good $2$ is $1/p_2$ at each bundle. Let $\beta=(1-\alpha)>0$. Then the marginal utility from $x_1$ is $\alpha 1/x^\beta$. For $x_1$ small enough, $\alpha 1/(x^\beta p_1)$ is larger than $1/p_2$.  

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### Answer by username_1 on 2020-12-14. Score: 2

An interesting issue arose while exchanging comments related to @MichaelGreineckeranswer. So let's explore this formally

> Claim: There is no boundary *solution* with
> $x\_1 = 0$.

Budget exhaustion holds, so with $x\_1 = 0$ we will have $U=x\_2 = w/p\_2$. So in an attempt to disprove the claim, we examine under which conditions the following inequality will hold:

$$\frac{w}{p\_2} > x\_1^{\alpha} + x\_2,\quad p\_1x\_1 + p\_2x\_2 = w.$$

Using the budget constraint to solve for $x\_2$ we can rewrite this

$$\frac{w}{p\_2} > x\_1^{\alpha} + \frac{w-p\_1x\_1}{p\_2}\;\implies\; \frac{p\_1}{p\_2}x\_1 > x\_1^{\alpha} \implies x\_1^{1-\alpha} > \frac{p\_2}{p\_1}$$

$$ \implies x\_1 > \left(\frac{p\_2}{p\_1}\right)^{1/(1-\alpha)} \tag{1}$$

So we have obtained: The utility from consuming $x\_2$ only, will be greater than the utility from any consumption plan for which the consumption bundle includes a quantity of $x\_1$ *higher* than the right-hand-side of $(1)$.

The crucial word her is "higher" -because it implies that bundles that include strictly positive amounts of both $(x\_1, x\_2)$ and as long as the amount of $x\_1$ is *lower* than the right-hand-side of $(1)$, are superior in terms of utility than consuming $x\_2$ only.

So @MichaelGreinecker is right: the solution set does not include a bundle with zero $x\_1$.


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