How to compute the Marshallian demand for this specific utility function

Question

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

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?

Answer 1

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 $x_3=0$.

Answer 2

$x_2$ 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.

Answer 3

Given the utility maximisation problem \begin{eqnarray*} \max_{(x_1,x_2,x_3)\in\mathbb{R}^3_+} & x_1^\alpha(x_2+x_3)^{1-\alpha} \\ \text{s.t. } & p_1x_1+p_2x_2+p_3x_3\leq M \end{eqnarray*} where $\alpha\in (0,1)$, $p_1>0$, $p_2>0$, $p_3>0$ and $M\geq 0$ are given. Solving it we get \begin{eqnarray*}(x_1^d,x_2^d,x_3^d)(p_1,p_2,p_3,M)\in\begin{cases} \left\{\left(\dfrac{\alpha M}{p_1},\dfrac{(1-\alpha) M}{p_2},0\right)\right\} & \text{if } p_2<p_3 \\ \left\{\left(\dfrac{\alpha M}{p_1},0,\dfrac{(1-\alpha) M}{p_3}\right)\right\} & \text{if } p_2>p_3 \\ \left\{\left(\dfrac{\alpha M}{p_1},\dfrac{\theta(1-\alpha) M}{p_2},\dfrac{(1-\theta)(1-\alpha) M}{p_3}\right)|\theta\in [0,1]\right\} & \text{if } p_2=p_3 \end{cases}\end{eqnarray*}

Answer 4

This is a typical case solvable by two-stage modeling.

As you notice the first stage is Cobb-Douglas

$$u(x_1,z) = x_1^\alpha z^{1-\alpha},$$

having solved this problem before you can probably write up the Marshall demand

$$x_1^\star = \frac{\alpha M}{p_1} \wedge z^\star=\frac{(1-\alpha)M}{p_z}$$

The second stage maximize $z = x_2+ x_3$ subject to $p_2x_2+p_3x_3\leq(1-\alpha)M$.

Again knowing the solution for perfect substitutes you know that $p_z=\min\{p_2,p_3\}$ and that demand is everything spent on the cheapest good and in the case of equality the some share of $(1-\alpha)M$ is spend on good 2 the rest on good 3 ... as Amit has written ....

My point is just that you can see it as a two-stage optimization problem.