# 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

$$$$\dfrac{\partial \mathcal{L}}{\partial x_1} = \alpha x_1 ^{\alpha -1} - \lambda p_1 + \mu_1=0,\\ \dfrac{\partial \mathcal{L}}{\partial x_2} = 1 - \lambda p_2 + \mu_2 =0,\\ p_1x_1 + p_2x_2 \leq w,\\ x_1\geq 0,\\ x_2 \geq 0,\\ \lambda(p_1x_1 + p_2x_2-w)=0,\\ \mu_1x_1 = 0,\\ \mu_2x_2=0,\\ \lambda \geq 0, \quad \mu_1 \geq 0, \quad \mu_2 \geq 0.\\$$$$

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

$$$$\alpha x_1 ^{\alpha -1} \leq \lambda p_1 ,\\ 1 = \lambda p_2,\\ x_1 = 0,\\ x_2 \geq 0.$$$$

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.

• Why your optimal candidate includes $p_1$ and not $p_2$? Dec 13 '20 at 18:33
• 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. Dec 13 '20 at 21:42
• @AlecosPapadopoulos it is a typo. Thanks for pointing it out. Dec 13 '20 at 21:45
• @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? Dec 13 '20 at 21:49
• It tells you that x1=0 x2>0 is not possible. Just like Greinecker says. But you still have chek x2=0 x1>0. Dec 13 '20 at 21:53

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 $$x_1$$.

• @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 $x_2$. Dec 13 '20 at 21:58
• How should I formalize (mathematically) the nonexistence of boundary solutions with $x_1=0$ (when not in extreme cases)? Dec 13 '20 at 22:07
• @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 $1/p_2$. Dec 13 '20 at 22:47

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 $$x_1$$.