<p>$x_2$ 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</p>

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

<p>This is the standard Cobb-Douglas function you already know how to solve.</p>

<p>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.</p>

CC BY-SA 3.0
2016-10-15T17:19:33.500Z
You provide a much deeper intuition of the mechanism here. Thanks!
Eric Chen
CC BY-SA 3.0
2016-10-03T19:27:40.333Z
2016-10-03T19:27:40.333Z
Ubiquitous
<p>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 $x_3=0$.</p>

CC BY-SA 3.0
2016-10-01T21:22:09.537Z
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.
Eric Chen
CC BY-SA 3.0
2016-10-02T01:07:00.263Z
@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 $x_1+x_2=C$.
Andrew
CC BY-SA 3.0
2016-10-01T18:06:10.557Z
2016-10-01T18:06:10.557Z
Andrew
<p>Given the utility maximisation problem
<span class="math-container">\begin{eqnarray*} \max_{(x_1,x_2,x_3)\in\mathbb{R}^3_+} &amp; x_1^\alpha(x_2+x_3)^{1-\alpha} \\ \text{s.t. } &amp; p_1x_1+p_2x_2+p_3x_3\leq M \end{eqnarray*}</span>
where <span class="math-container">$\alpha\in (0,1)$</span>, <span class="math-container">$p_1&gt;0$</span>, <span class="math-container">$p_2&gt;0$</span>, <span class="math-container">$p_3&gt;0$</span> and <span class="math-container">$M\geq 0$</span> are given. Solving it we get
<span class="math-container">\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\} &amp; \text{if } p_2&lt;p_3 \\ \left\{\left(\dfrac{\alpha M}{p_1},0,\dfrac{(1-\alpha) M}{p_3}\right)\right\} &amp; \text{if } p_2&gt;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\} &amp; \text{if } p_2=p_3 \end{cases}\end{eqnarray*}</span></p>

CC BY-SA 4.0
2024-08-21T05:57:06.757Z
2024-08-21T05:57:06.757Z
Amit
<p>This is a typical case solvable by two-stage modeling.</p>
<p>As you notice the first stage is Cobb-Douglas</p>
<p><span class="math-container">$$u(x_1,z) = x_1^\alpha z^{1-\alpha},$$</span></p>
<p>having solved this problem before you can probably write up the Marshall demand</p>
<p><span class="math-container">$$x_1^\star = \frac{\alpha M}{p_1} \wedge z^\star=\frac{(1-\alpha)M}{p_z}$$</span></p>
<p>The second stage maximize <span class="math-container">$z = x_2+ x_3$</span> subject to <span class="math-container">$p_2x_2+p_3x_3\leq(1-\alpha)M$</span>.</p>
<p>Again knowing the solution for perfect substitutes you know that <span class="math-container">$p_z=\min\{p_2,p_3\}$</span> and that demand is everything spent on the cheapest good and in the case of equality the some share of <span class="math-container">$(1-\alpha)M$</span> is spend on good 2 the rest on good 3 ... as Amit has written ....</p>
<p>My point is just that you can see it as a two-stage optimization problem.</p>

CC BY-SA 4.0
2024-08-21T21:11:01.500Z
2024-08-21T21:11:01.500Z
Jesper Hybel
<p>$$u(x) = x_1^\alpha \cdot (x_2+x_3)^{1-\alpha}, \text{ with } \alpha \in(0,1)$$</p>

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

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

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

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

<p>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. </p>

<p>Any one can help?</p>

CC BY-SA 4.0
2022-05-27T05:30:12.433Z
You need to use Kuhn Tucker method to solve this problem because of possibility of a corner solution.
Amit
<utility>
How to compute the Marshallian demand for this specific utility function
Eric Chen
# How to compute the Marshallian demand for this specific utility function

(Asked by: username_0 on 2016-10-01)

$$u(x) = x\_1^\alpha \cdot (x\_2+x\_3)^{1-\alpha}, \text{ with } \alpha \in(0,1)$$

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?

**Tags:** utility

> **Comments:**
> **username_1**: You need to use Kuhn Tucker method to solve this problem because of possibility of a corner solution.  

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## 4 answers

### Answer by username_2 on 2016-10-03. Score: 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.

> **Comments:**
> **username_0**: You provide a much deeper intuition of the mechanism here. Thanks!  

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### Answer by username_3 on 2016-10-01. Score: 2

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 $x\_3=0$.

> **Comments:**
> **username_0**: 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.  
> **username_3**: @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 $x_1+x_2=C$.  

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### Answer by username_1 on 2024-08-21. Score: 1

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\_2p\_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\*}


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### Answer by username_4 on 2024-08-21. Score: 0

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 username_1 has written ....

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


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