Boundary solutions in the Utility Maximization Problem 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 $x_2&gt;0$, 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 $p_1,p_2 &gt; 0$, $\lambda &gt; 0$. 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.
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 $x_1$.