$$u(x) = x_1^\alpha \cdot (x_2+x_3)^{1-\alpha}, \text{ with } \alpha \in(0,1)$$

I tried to set up the Lagrangian and it turns out $\lambda$ have no solution unless $p_1 = p_2$.

$$L = x_1^\alpha \cdot (x_2+x_3)^{1-\alpha} -\lambda(x_1 p_1 + x_2 p_2 + x_3 p_3 -1)$$

$$\frac{\partial L}{\partial x_2} = (1-\alpha)x_1^\alpha(x_2+x_3)^{-\alpha} -\lambda p_2 = 0$$

$$\frac{\partial L}{\partial x_3} = (1-\alpha)x_1^\alpha(x_2+x_3)^{-\alpha} -\lambda p_3 = 0$$

I noticed this function is similar to C-D utility function. So I'm wondering if I can combine $x_{3}$ and $x_{2}$ as one good and then apply C-S utility function. But this method seems unreliable to me.

Any one can help?


It means that the level surfaces of $u$ don't touch the surface $x_1p_1+x_2p_2+x_3p_3=1$ unless $p_2=p_3$. But, since $x_i\ge0$, it follows that the admissible domain of $x_i$ is compact and the maximum is still achieved somewhere. Namely, in some boundary point of the admissible domain $-$ for $x_2=0$ or $x_3=0$.

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  • $\begingroup$ My intuition also tells me that, either $x_{2}$ or $x_{3}$ would be 0. I'm still working on a more rigorous proof for that. $\endgroup$ – Eric Chen Oct 1 '16 at 21:22
  • $\begingroup$ @EricChen What I've said is quite rigorous. May be it would be helpful to consider beforehand a case of two variables with $u(x)=x_1+x_2$ and the line $p_1x_1+x_2p_2=1$. And solve it geometrically drawing lines of level lines $x_1+x_2=C$. $\endgroup$ – Andrew Oct 2 '16 at 1:07

$x_2$ and $x_3$ are perfect substitutes. Indeed, I can always take away one unit of $x_2$ and give one extra unit of $x_3$ and utility will remain exactly unchanged. This implies that the consumer will only ever buy whichever of $x_2$ or $x_3$ is cheapest. In other words, if $p_2\leq p_3$ then there is no loss in generality from setting $x_3=0$. The problem then becomes

$$\max_{x_1,x_2}x_1^a x_2^{1-a}$$ subject to $$p_1 x_1+p_2 x_2=M$$

This is the standard Cobb-Douglas function you already know how to solve.

This is actually quite a neat question because it forces us to think about the mechanism of what the consumer is doing when he maximises utility, rather than just mindlessly setting up the optimisation problem.

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  • $\begingroup$ You provide a much deeper intuition of the mechanism here. Thanks! $\endgroup$ – Eric Chen Oct 15 '16 at 17:19

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