<p>No, you can't. The Karush-Kuhn-Tucker theorem is for functions defined on <span class="math-container">$\mathbb{R}^n$</span> or at least an open subset thereof and requires the function to be maximized to be differentiable on the domain. That the solution has to lie in the nonnegative orthant is not something that can be represented by inequality constraints here, since the utility function is not defined on a larger open domain on which it is differentiable. Indeed, the function <span class="math-container">$(x_1,x_2)\mapsto \sqrt{x_1x_2}$</span> is not differentiable at points where <span class="math-container">$x_1=0$</span> or <span class="math-container">$x_2=0$</span>. To see this, note that if the function is <a href="https://en.wikipedia.org/wiki/Differentiable_function#Differentiability_in_higher_dimensions" rel="nofollow noreferrer">differentiable</a> at, for example, the point <span class="math-container">$(x_1,x_2)=(0,1)$</span>, then there would exists a vector <span class="math-container">$v=(v_1,v_2)$</span> such that
<span class="math-container">$$\lim_{h\to 0}\frac{\|\sqrt{(x_1+h_1)(x_2+h_2)}-\sqrt{x_1x_2}-v_1x_1-v_2x_2 \|}{\|h\|}=0$$</span>
<span class="math-container">$$\lim_{h\to 0}\frac{\|\sqrt{h_1+h_1h_2}-v_2 \|}{\|h\|}=0.$$</span>
But, for example, taking for <span class="math-container">$h=(0,1/n)$</span>
<span class="math-container">$$\lim_{n\to \infty}\frac{\|\sqrt{0+0\cdot 1/n}-v_2 \|}{1/n}=\lim_{n\to\infty }n\|-v_2\|,$$</span>
which is only <span class="math-container">$0$</span> if <span class="math-container">$v_2=0$</span>. So assume <span class="math-container">$v_2=0$</span> and take <span class="math-container">$h=(1/n,1/n)$</span> instead,
you get
<span class="math-container">$$\liminf_{n\to \infty}\frac{\|\sqrt{1/n+1/n^2}\|}{\sqrt 2~ 1/n}\geq 1/\sqrt 2.$$</span>
No derivative exists at <span class="math-container">$(0,1)$</span>.</p>
<p>Misuse of the Karush-Kuhn-Tucker theorem is very prevalent, even in otherwise very rigorous textbooks. The appendix of the following paper lists several examples of textbooks that apply the KKT-theorem to problems where it can't:</p>
<blockquote>
<p>Erik J. Balder, &quot;<a href="https://webspace.science.uu.nl/%7Ebalde101/giffenvol.pdf#page=15" rel="nofollow noreferrer">Exact and useful optimization methods for
microeconomics</a>&quot; in: New Insights into the Theory of Giffen Goods
(W. Heijman and P. Mouche, eds.), Lecture Notes in Economics and
Mathematical Systems 655, Springer, 2012, pp. 21-38.</p>
</blockquote>

CC BY-SA 4.0
2022-05-28T05:04:08.727Z
Thanks Michael. Regarding your comment about the misuse, let me clarify that the conditions in the theorem are sufficient conditions. That means it is possible that in conditions other than the ones that are mentioned in the theorem, the KKT conditions can still be used. For example, in situations where solutions to the optimization problems does not lie on the boundaries (where the function is not differentiable) we can use KKT conditions to find the solutions even when the function is not differentiable at these points as long as it is  continuous at those points.
Amit
CC BY-SA 4.0
2022-05-28T05:11:23.917Z
Let me know if you want me to prove this. I can provide you the proof. But yes, your point about this particular problem is well taken (since the solution lies on the boundary). And I have presented an alternative way to do this using KKT method in my another answer.
Amit
CC BY-SA 4.0
2022-05-28T07:44:20.313Z
If we know that the solution does not lie on the boundary, we can of course use $\mathbb{R}_{++}$ as the domain. This is the case of "neoclassical preferences" as definedAliprantis  Brown, Burkinshaw, which is also mentioned in Balder's paper.
Michael Greinecker
CC BY-SA 4.0
2022-05-28T07:45:42.277Z
That’s right. That’s the proof 
Amit
CC BY-SA 4.0
2022-06-06T22:32:18.047Z
But on the open convex set $\{(x_1,x_2,x_3,x_4)\in\mathbb{R}^4| x_i > 0\}$ the function IS differentiable.
Davius
CC BY-SA 4.0
2022-06-07T00:17:42.590Z
@Davius That's what the previous comment said.
Michael Greinecker
CC BY-SA 4.0
2022-05-27T08:38:41.827Z
2022-05-27T14:31:19.083Z
2022-05-27T14:31:19.083Z
Michael Greinecker
<p>Why do you want to guess that some <span class="math-container">$x$</span> is zero? What is the problem of having all four goods strictly positive at the solution?  </p>

<p>In fact, due to the presence of the square root in the first order conditions, the solution looks straightforward, since the inequality cannot hold (and assuming exhaustion of the budget).  </p>

<hr>

<p><strong>ADDENDUM</strong><br>
Given the f.o.c's provided by the OP, all <span class="math-container">$x$</span>'s have to be strictly positive at the solution, since otherwise we have division by zero and/or the undetermined form <span class="math-container">$0/0$</span>.  </p>

<p>This in turn implies that <span class="math-container">$\partial L/\partial x_i = 0$</span> at the solution, which now becomes easy to compute.</p>

CC BY-SA 4.0
2018-10-30T05:39:59.547Z
It's because the solutions are (i) if $p_1p_2 < p_3 p_4$, then $x_1 = \frac{M}{2p_1}$, $x_2 = \frac{M}{2p_2}$, $x_3 =x_4 =0$; (ii) if $p_1p_2 > p_3 p_4$, then $x_3 = \frac{M}{2p_3}$, $x_4 = \frac{M}{2p_4}$, $x_1 =x_2 =0$; and (iii) with an interior solution, when $p_1p_2 = p_3 p_4$.
Yun
CC BY-SA 4.0
2018-10-29T17:51:31.793Z
2018-11-29T02:53:47.180Z
2018-11-29T02:53:47.180Z
Alecos Papadopoulos
<p><span class="math-container">\begin{eqnarray*}\max_{x_1,x_2, x_3, x_4} &amp; \sqrt{x_1x_2} + \sqrt{x_3x_4} \\ \text{s.t.} &amp; \ p_1x_1 + p_2x_2+p_3x_3 +p_4x_4 \leq w \\ \text{and } &amp; \ x_1, x_2, x_3, x_4 \geq 0\end{eqnarray*}</span></p>
<p>One way to solve this problem is using Kuhn-Tucker method.</p>
<p>Here is another way to solve this problem:</p>
<p>Consider the following two-stage problem:</p>
<ul>
<li>In stage 1, allocate <span class="math-container">$w_{12}\geq 0, w_{34}\geq 0$</span> out of the entire income <span class="math-container">$w$</span> (such that <span class="math-container">$w_{12} + w_{34}\leq w$</span>) to spend on bundles <span class="math-container">$(x_1, x_2)$</span> and <span class="math-container">$(x_3, x_4)$</span> respectively, in the best possible way.</li>
<li>In stage 2, spend the allocated amount in the best possible way.</li>
</ul>
<p>Formally, this is how we can define the problem:
<span class="math-container">\begin{eqnarray*}\max_{w_{12}\geq 0, \ w_{34} \geq 0, \ w_{12} +w_{34} \leq w} \left[\max_{x_1\geq 0, \ x_2 \geq 0, \ p_1x_1 + p_2x_2 \leq w_{12}}\sqrt{x_1x_2} + \max_{x_3\geq 0, \ x_4 \geq 0, \ p_3x_3 + p_4x_4 \leq w_{34}}\sqrt{x_3x_4} \right] \end{eqnarray*}</span></p>
<p>It can be easily shown that the solution set to this problem will be same as the solution set of the problem we want to solve. Let me know if you want to see the proof.</p>
<p>We can now solve this by backward induction:</p>
<p>Solution to the problems inside the parenthesis is:</p>
<p><span class="math-container">\begin{eqnarray*} (x_1^i, x_2^i)(p_1, p_2, w_{12}) &amp; = &amp; \left(\frac{w_{12}}{2p_1},\frac{w_{12}}{2p_2} \right) \\ (x_3^i, x_4^i)(p_3, p_4, w_{34}) &amp; = &amp; \left(\frac{w_{34}}{2p_3},\frac{w_{34}}{2p_4} \right)\end{eqnarray*}</span></p>
<p>Using the solution above, we can re-write the problem as:
<span class="math-container">\begin{eqnarray*}\max_{w_{12}, w_{34}} &amp; \frac{w_{12}}{2\sqrt{p_1p_2}} + \frac{w_{34}}{2\sqrt{p_3p_4}} \\ \text{s.t.} &amp; \ \ w_{12}\geq 0, \ w_{34} \geq 0, \ w_{12} +w_{34} \leq w  \end{eqnarray*}</span></p>
<p>Now this is a standard perfect substitutes case, and we get</p>
<p><span class="math-container">\begin{eqnarray*} (w_{12},w_{34})(p_1, p_2, p_3, p_4, w) \in \begin{cases} \left\{\left(w, 0\right)\right\} &amp; \text{if } p_1p_2 &lt; p_3p_4 \\ \left\{\left(0,w\right)\right\} &amp; \text{if } p_1p_2 &gt; p_3p_4 \\ \left\{\left(m,n\right)\in\mathbb{R}^2_+| m+n=w\right\} &amp; \text{if } p_1p_2 = p_3p_4\end{cases} \end{eqnarray*}</span></p>
<p>Now we can substitute this back in <span class="math-container">$(x_1^i, x_2^i)$</span> and <span class="math-container">$(x_3^i, x_4^i)$</span> to get the demands:
<span class="math-container">\begin{eqnarray*} (x_1^d,x_2^d, x_3^d, x_4^d)(p_1, p_2, p_3, p_4, w) \in \begin{cases} \left\{\left(\frac{w}{2p_1},\frac{w}{2p_2}, 0, 0\right)\right\} &amp; \text{if } p_1p_2 &lt; p_3p_4 \\ \left\{\left(0,0,\frac{w}{2p_3},\frac{w}{2p_4}\right)\right\} &amp; \text{if } p_1p_2 &gt; p_3p_4 \\ \left\{\left(\frac{m}{2p_1},\frac{m}{2p_2},\frac{n}{2p_3},\frac{n}{2p_4}\right)\in\mathbb{R}^4_+| m+n=w\right\} &amp; \text{if } p_1p_2 = p_3p_4\end{cases} \end{eqnarray*}</span></p>

CC BY-SA 4.0
2022-05-27T04:57:50.930Z
2022-05-27T05:04:18.100Z
2022-05-27T05:04:18.100Z
Amit
<p>As Michael pointed out that we cannot use Kuhn-Tucker conditions here because the objective function is not differentiable everywhere (at the boundary). Here is what we can do in light of the above:</p>
<p><strong>Theorem</strong>. Let <span class="math-container">$u:\mathbb{R}^n_+\rightarrow\mathbb{R}$</span> be continuously differentiable on <span class="math-container">$\mathbb{R}^n_{++}$</span>, increasing and continuous everywhere and suppose that <span class="math-container">$u$</span> is concave and prices <span class="math-container">$p_i&gt;0 \text{ for each }  i \in \{1, 2\ldots, n\}$</span>and <span class="math-container">$w&gt;0$</span> then <span class="math-container">$(x_1^*, x_2^*, \ldots, x_n^*)$</span> solves the problem
<span class="math-container">\begin{eqnarray*} \max_{x_1, x_2, \ldots, x_n} &amp; u(x_1, x_2,\ldots, x_n) \\ \text{s.t.} &amp; \ \sum_{i=1}^{n} p_ix_i \leq  w  \\ \text{and } &amp; x_i \geq \epsilon_i &amp; \text{ for each }  i \in \{1, 2\ldots, n\}\end{eqnarray*}</span> for all <span class="math-container">$\epsilon_i&gt; 0$</span> (and sufficiently small so that the constraint set is non-empty).</p>
<p><em>if and only if</em> it satisfies the Kuhn-Tucker conditions.</p>
<p>Of course, the above is true by the Kuhn-Tucker Theorem.</p>
<p><strong>Proposition</strong>: This is how we can use the Kuhn-Tucker conditions to solve the problem we are interested in i.e. when <span class="math-container">$\epsilon_i=0$</span>:</p>
<p>Since <span class="math-container">$p_i&gt;0 \text{ for each }  i \in \{1, 2\ldots, n\}$</span>, constraint set is compact and convex, and given that the utility function is concave and continuous, the solution set is non-empty, convex and compact. And the solution set to the Kuhn-Tucker conditions is same as the solution set of the optimization problem (by the above theorem). Since the solution set to the optimization problem (including the case where <span class="math-container">$\epsilon_i$</span>s are possibly <span class="math-container">$0$</span>) is a upper hemi-continuous correspondence in <span class="math-container">$\epsilon = (\epsilon_1, \epsilon_2, \ldots, \epsilon_n)$</span> (under our assumptions), it has a closed graph. So, we can just solve the problem above using the Kuhn-Tucker conditions for positive <span class="math-container">$\epsilon_i$</span>s and then find the limit points of the graph of the solutions. Since graph is closed, set of points at <span class="math-container">$(\epsilon_1, \epsilon_2, \ldots, \epsilon_n)= 0$</span> in the closure of the graph will be the solution of the desired optimization problem and this is how we can now use the Kuhn-Tucker conditions to solve the problem. This approach may not give all the solutions to the desired optimization problem but will definitely give a non-empty subset of the solutions.*</p>
<p>*In the problem with <span class="math-container">$u= \sqrt{x_1x_2} +\sqrt{x_3x_4}$</span>, it'll give all the solutions, and right now I cannot think of a case where this approach will not give all the solutions.</p>

CC BY-SA 4.0
2022-05-28T01:27:04.150Z
2022-05-28T06:12:29.173Z
2022-05-28T06:12:29.173Z
Amit
<p>Consider </p>

<p><span class="math-container">$\max_{x_1, x_2, x_3, x_4} u(x) = \sqrt{x_1 x_2} + \sqrt{x_3 x_4}$</span></p>

<p>s.t. <span class="math-container">$\; p_1x_1 + p_2x_2 + p_3x_3 + p_4x_4 \le w$</span></p>

<p>I know we can solve the max problem through separately considering case(i): <span class="math-container">$x_1, x_2 &gt; 0$</span> and <span class="math-container">$x_3 = x_4 = 0$</span>; and case (ii) <span class="math-container">$x_1 = x_2 = 0$</span> and <span class="math-container">$x_3, x_4 &gt; 0$</span>.</p>

<p>But is it possible to solve the whole optimization problem through the Kuhn–Tucker method?</p>

<p>We can write down the Lagrangian <span class="math-container">$L(x,\lambda)=\sqrt{x_1x_2}+\sqrt{x_3x_4}+\lambda(w-p_1x_1-p_2x_2-p_3x_3-p_4x_4)$</span></p>

<p>with the complementary slackness conditions:</p>

<p><span class="math-container">$\frac{\partial L}{\partial x_1}=\frac{1}{2}\sqrt{\frac{x_2}{x_1}}-\lambda p_1 \le 0,\quad x_1 \ge 0, \quad \text{and}\quad  x_1 \frac{\partial L}{\partial x_1}=0$</span>.</p>

<p><span class="math-container">$\frac{\partial L}{\partial x_2}=\frac{1}{2}\sqrt{\frac{x_1}{x_2}}-\lambda p_2 \le 0,\quad x_2 \ge 0, \quad \text{and}\quad  x_2 \frac{\partial L}{\partial x_2}=0$</span>.</p>

<p><span class="math-container">$\frac{\partial L}{\partial x_3}=\frac{1}{2}\sqrt{\frac{x_4}{x_3}}-\lambda p_3 \le 0,\quad x_3 \ge 0, \quad \text{and}\quad  x_3 \frac{\partial L}{\partial x_3}=0$</span>.</p>

<p><span class="math-container">$\frac{\partial L}{\partial x_4}=\frac{1}{2}\sqrt{\frac{x_3}{x_4}}-\lambda p_4 \le 0,\quad x_4 \ge 0, \quad \text{and}\quad  x_4 \frac{\partial L}{\partial x_4}=0$</span>.</p>

<p>However, when guessing say, <span class="math-container">$x_1 = 0$</span> and <span class="math-container">$x_2,x_3,x_4&gt;0$</span>, <span class="math-container">$\lim_{x_1 \to 0}\frac{\partial L}{\partial x_1} \to \infty$</span>, which does not satisfy the complementary slackness <span class="math-container">$x_1 \frac{\partial L}{\partial x_1} = 0$</span>.</p>

<p>I do not know whether we can use the Kuhn–Tucker method to solve this optimization problem? And if not, what are the reasons?</p>

CC BY-SA 4.0
2018-10-30T01:59:57.073Z
Here's a good trick for any microeconomic optimization problem: if the agent's preferences are locally non-satiated, then Walras' Law will hold. That is, $p x = w$. Also note that monotonic preferences are locally non-satiated, and as your objective function is weakly increasing in each of its arguments, you have weakly monotonic preferences.
NBm424
<microeconomics><optimization>
Can $u(x) = \sqrt{x_1 x_2} + \sqrt{x_3 x_4}$ be solved by Kuhn–Tucker conditions?
Yun
# Can $u(x) = \sqrt{x_1 x_2} + \sqrt{x_3 x_4}$ be solved by Kuhn–Tucker conditions?

(Asked by: username_0 on 2018-10-29)

Consider

$\max\_{x\_1, x\_2, x\_3, x\_4} u(x) = \sqrt{x\_1 x\_2} + \sqrt{x\_3 x\_4}$

s.t. $\; p\_1x\_1 + p\_2x\_2 + p\_3x\_3 + p\_4x\_4 \le w$

I know we can solve the max problem through separately considering case(i): $x\_1, x\_2 > 0$ and $x\_3 = x\_4 = 0$; and case (ii) $x\_1 = x\_2 = 0$ and $x\_3, x\_4 > 0$.

But is it possible to solve the whole optimization problem through the Kuhn–Tucker method?

We can write down the Lagrangian $L(x,\lambda)=\sqrt{x\_1x\_2}+\sqrt{x\_3x\_4}+\lambda(w-p\_1x\_1-p\_2x\_2-p\_3x\_3-p\_4x\_4)$

with the complementary slackness conditions:

$\frac{\partial L}{\partial x\_1}=\frac{1}{2}\sqrt{\frac{x\_2}{x\_1}}-\lambda p\_1 \le 0,\quad x\_1 \ge 0, \quad \text{and}\quad x\_1 \frac{\partial L}{\partial x\_1}=0$.

$\frac{\partial L}{\partial x\_2}=\frac{1}{2}\sqrt{\frac{x\_1}{x\_2}}-\lambda p\_2 \le 0,\quad x\_2 \ge 0, \quad \text{and}\quad x\_2 \frac{\partial L}{\partial x\_2}=0$.

$\frac{\partial L}{\partial x\_3}=\frac{1}{2}\sqrt{\frac{x\_4}{x\_3}}-\lambda p\_3 \le 0,\quad x\_3 \ge 0, \quad \text{and}\quad x\_3 \frac{\partial L}{\partial x\_3}=0$.

$\frac{\partial L}{\partial x\_4}=\frac{1}{2}\sqrt{\frac{x\_3}{x\_4}}-\lambda p\_4 \le 0,\quad x\_4 \ge 0, \quad \text{and}\quad x\_4 \frac{\partial L}{\partial x\_4}=0$.

However, when guessing say, $x\_1 = 0$ and $x\_2,x\_3,x\_4>0$, $\lim\_{x\_1 \to 0}\frac{\partial L}{\partial x\_1} \to \infty$, which does not satisfy the complementary slackness $x\_1 \frac{\partial L}{\partial x\_1} = 0$.

I do not know whether we can use the Kuhn–Tucker method to solve this optimization problem? And if not, what are the reasons?

**Tags:** microeconomics, optimization
> **Comments:**
> **username_1**: Here's a good trick for any microeconomic optimization problem: if the agent's preferences are locally non-satiated, then Walras' Law will hold. That is, $p x = w$. Also note that monotonic preferences are locally non-satiated, and as your objective function is weakly increasing in each of its arguments, you have weakly monotonic preferences.  

---

## 4 answers

### Answer by username_2 on 2022-05-27. Score: 4

No, you can't. The Karush-Kuhn-Tucker theorem is for functions defined on $\mathbb{R}^n$ or at least an open subset thereof and requires the function to be maximized to be differentiable on the domain. That the solution has to lie in the nonnegative orthant is not something that can be represented by inequality constraints here, since the utility function is not defined on a larger open domain on which it is differentiable. Indeed, the function $(x\_1,x\_2)\mapsto \sqrt{x\_1x\_2}$ is not differentiable at points where $x\_1=0$ or $x\_2=0$. To see this, note that if the function is [differentiable](https://en.wikipedia.org/wiki/Differentiable_function#Differentiability_in_higher_dimensions) at, for example, the point $(x\_1,x\_2)=(0,1)$, then there would exists a vector $v=(v\_1,v\_2)$ such that
$$\lim\_{h\to 0}\frac{\|\sqrt{(x\_1+h\_1)(x\_2+h\_2)}-\sqrt{x\_1x\_2}-v\_1x\_1-v\_2x\_2 \|}{\|h\|}=0$$
$$\lim\_{h\to 0}\frac{\|\sqrt{h\_1+h\_1h\_2}-v\_2 \|}{\|h\|}=0.$$
But, for example, taking for $h=(0,1/n)$
$$\lim\_{n\to \infty}\frac{\|\sqrt{0+0\cdot 1/n}-v\_2 \|}{1/n}=\lim\_{n\to\infty }n\|-v\_2\|,$$
which is only $0$ if $v\_2=0$. So assume $v\_2=0$ and take $h=(1/n,1/n)$ instead,
you get
$$\liminf\_{n\to \infty}\frac{\|\sqrt{1/n+1/n^2}\|}{\sqrt 2~ 1/n}\geq 1/\sqrt 2.$$
No derivative exists at $(0,1)$.

Misuse of the Karush-Kuhn-Tucker theorem is very prevalent, even in otherwise very rigorous textbooks. The appendix of the following paper lists several examples of textbooks that apply the KKT-theorem to problems where it can't:

> Erik J. Balder, "[Exact and useful optimization methods for
> microeconomics](https://webspace.science.uu.nl/%7Ebalde101/giffenvol.pdf#page=15)" in: New Insights into the Theory of Giffen Goods
> (W. Heijman and P. Mouche, eds.), Lecture Notes in Economics and
> Mathematical Systems 655, Springer, 2012, pp. 21-38.

> **Comments:**
> **username_3**: Thanks Michael. Regarding your comment about the misuse, let me clarify that the conditions in the theorem are sufficient conditions. That means it is possible that in conditions other than the ones that are mentioned in the theorem, the KKT conditions can still be used. For example, in situations where solutions to the optimization problems does not lie on the boundaries (where the function is not differentiable) we can use KKT conditions to find the solutions even when the function is not differentiable at these points as long as it is  continuous at those points.  
> **username_3**: Let me know if you want me to prove this. I can provide you the proof. But yes, your point about this particular problem is well taken (since the solution lies on the boundary). And I have presented an alternative way to do this using KKT method in my another answer.  
> **username_2**: If we know that the solution does not lie on the boundary, we can of course use $\mathbb{R}_{++}$ as the domain. This is the case of "neoclassical preferences" as definedAliprantis  Brown, Burkinshaw, which is also mentioned in Balder's paper.  
> **username_3**: That’s right. That’s the proof   
> **username_4**: But on the open convex set $\{(x_1,x_2,x_3,x_4)\in\mathbb{R}^4| x_i > 0\}$ the function IS differentiable.  
> **username_2**: @Davius That's what the previous comment said.  

---

### Answer by username_5 on 2018-10-29. Score: 3

Why do you want to guess that some $x$ is zero? What is the problem of having all four goods strictly positive at the solution?

In fact, due to the presence of the square root in the first order conditions, the solution looks straightforward, since the inequality cannot hold (and assuming exhaustion of the budget).

---

**ADDENDUM**  
Given the f.o.c's provided by the OP, all $x$'s have to be strictly positive at the solution, since otherwise we have division by zero and/or the undetermined form $0/0$.

This in turn implies that $\partial L/\partial x\_i = 0$ at the solution, which now becomes easy to compute.

> **Comments:**
> **username_0**: It's because the solutions are (i) if $p_1p_2 < p_3 p_4$, then $x_1 = \frac{M}{2p_1}$, $x_2 = \frac{M}{2p_2}$, $x_3 =x_4 =0$; (ii) if $p_1p_2 > p_3 p_4$, then $x_3 = \frac{M}{2p_3}$, $x_4 = \frac{M}{2p_4}$, $x_1 =x_2 =0$; and (iii) with an interior solution, when $p_1p_2 = p_3 p_4$.  

---

### Answer by username_3 on 2022-05-27. Score: 3

\begin{eqnarray\*}\max\_{x\_1,x\_2, x\_3, x\_4} & \sqrt{x\_1x\_2} + \sqrt{x\_3x\_4} \\ \text{s.t.} & \ p\_1x\_1 + p\_2x\_2+p\_3x\_3 +p\_4x\_4 \leq w \\ \text{and } & \ x\_1, x\_2, x\_3, x\_4 \geq 0\end{eqnarray\*}

One way to solve this problem is using Kuhn-Tucker method.

Here is another way to solve this problem:

Consider the following two-stage problem:

* In stage 1, allocate $w\_{12}\geq 0, w\_{34}\geq 0$ out of the entire income $w$ (such that $w\_{12} + w\_{34}\leq w$) to spend on bundles $(x\_1, x\_2)$ and $(x\_3, x\_4)$ respectively, in the best possible way.
* In stage 2, spend the allocated amount in the best possible way.

Formally, this is how we can define the problem:
\begin{eqnarray\*}\max\_{w\_{12}\geq 0, \ w\_{34} \geq 0, \ w\_{12} +w\_{34} \leq w} \left[\max\_{x\_1\geq 0, \ x\_2 \geq 0, \ p\_1x\_1 + p\_2x\_2 \leq w\_{12}}\sqrt{x\_1x\_2} + \max\_{x\_3\geq 0, \ x\_4 \geq 0, \ p\_3x\_3 + p\_4x\_4 \leq w\_{34}}\sqrt{x\_3x\_4} \right] \end{eqnarray\*}

It can be easily shown that the solution set to this problem will be same as the solution set of the problem we want to solve. Let me know if you want to see the proof.

We can now solve this by backward induction:

Solution to the problems inside the parenthesis is:

\begin{eqnarray\*} (x\_1^i, x\_2^i)(p\_1, p\_2, w\_{12}) & = & \left(\frac{w\_{12}}{2p\_1},\frac{w\_{12}}{2p\_2} \right) \\ (x\_3^i, x\_4^i)(p\_3, p\_4, w\_{34}) & = & \left(\frac{w\_{34}}{2p\_3},\frac{w\_{34}}{2p\_4} \right)\end{eqnarray\*}

Using the solution above, we can re-write the problem as:
\begin{eqnarray\*}\max\_{w\_{12}, w\_{34}} & \frac{w\_{12}}{2\sqrt{p\_1p\_2}} + \frac{w\_{34}}{2\sqrt{p\_3p\_4}} \\ \text{s.t.} & \ \ w\_{12}\geq 0, \ w\_{34} \geq 0, \ w\_{12} +w\_{34} \leq w \end{eqnarray\*}

Now this is a standard perfect substitutes case, and we get

\begin{eqnarray\*} (w\_{12},w\_{34})(p\_1, p\_2, p\_3, p\_4, w) \in \begin{cases} \left\{\left(w, 0\right)\right\} & \text{if } p\_1p\_2 < p\_3p\_4 \\ \left\{\left(0,w\right)\right\} & \text{if } p\_1p\_2 > p\_3p\_4 \\ \left\{\left(m,n\right)\in\mathbb{R}^2\_+| m+n=w\right\} & \text{if } p\_1p\_2 = p\_3p\_4\end{cases} \end{eqnarray\*}

Now we can substitute this back in $(x\_1^i, x\_2^i)$ and $(x\_3^i, x\_4^i)$ to get the demands:
\begin{eqnarray\*} (x\_1^d,x\_2^d, x\_3^d, x\_4^d)(p\_1, p\_2, p\_3, p\_4, w) \in \begin{cases} \left\{\left(\frac{w}{2p\_1},\frac{w}{2p\_2}, 0, 0\right)\right\} & \text{if } p\_1p\_2 < p\_3p\_4 \\ \left\{\left(0,0,\frac{w}{2p\_3},\frac{w}{2p\_4}\right)\right\} & \text{if } p\_1p\_2 > p\_3p\_4 \\ \left\{\left(\frac{m}{2p\_1},\frac{m}{2p\_2},\frac{n}{2p\_3},\frac{n}{2p\_4}\right)\in\mathbb{R}^4\_+| m+n=w\right\} & \text{if } p\_1p\_2 = p\_3p\_4\end{cases} \end{eqnarray\*}


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### Answer by username_3 on 2022-05-28. Score: 3

As Michael pointed out that we cannot use Kuhn-Tucker conditions here because the objective function is not differentiable everywhere (at the boundary). Here is what we can do in light of the above:

**Theorem**. Let $u:\mathbb{R}^n\_+\rightarrow\mathbb{R}$ be continuously differentiable on $\mathbb{R}^n\_{++}$, increasing and continuous everywhere and suppose that $u$ is concave and prices $p\_i>0 \text{ for each } i \in \{1, 2\ldots, n\}$and $w>0$ then $(x\_1^\*, x\_2^\*, \ldots, x\_n^\*)$ solves the problem
\begin{eqnarray\*} \max\_{x\_1, x\_2, \ldots, x\_n} & u(x\_1, x\_2,\ldots, x\_n) \\ \text{s.t.} & \ \sum\_{i=1}^{n} p\_ix\_i \leq w \\ \text{and } & x\_i \geq \epsilon\_i & \text{ for each } i \in \{1, 2\ldots, n\}\end{eqnarray\*} for all $\epsilon\_i> 0$ (and sufficiently small so that the constraint set is non-empty).

*if and only if* it satisfies the Kuhn-Tucker conditions.

Of course, the above is true by the Kuhn-Tucker Theorem.

**Proposition**: This is how we can use the Kuhn-Tucker conditions to solve the problem we are interested in i.e. when $\epsilon\_i=0$:

Since $p\_i>0 \text{ for each } i \in \{1, 2\ldots, n\}$, constraint set is compact and convex, and given that the utility function is concave and continuous, the solution set is non-empty, convex and compact. And the solution set to the Kuhn-Tucker conditions is same as the solution set of the optimization problem (by the above theorem). Since the solution set to the optimization problem (including the case where $\epsilon\_i$s are possibly $0$) is a upper hemi-continuous correspondence in $\epsilon = (\epsilon\_1, \epsilon\_2, \ldots, \epsilon\_n)$ (under our assumptions), it has a closed graph. So, we can just solve the problem above using the Kuhn-Tucker conditions for positive $\epsilon\_i$s and then find the limit points of the graph of the solutions. Since graph is closed, set of points at $(\epsilon\_1, \epsilon\_2, \ldots, \epsilon\_n)= 0$ in the closure of the graph will be the solution of the desired optimization problem and this is how we can now use the Kuhn-Tucker conditions to solve the problem. This approach may not give all the solutions to the desired optimization problem but will definitely give a non-empty subset of the solutions.\*

\*In the problem with $u= \sqrt{x\_1x\_2} +\sqrt{x\_3x\_4}$, it'll give all the solutions, and right now I cannot think of a case where this approach will not give all the solutions.


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