<p>This is the problem we want to solve:
<span class="math-container">\begin{eqnarray*} \max_{x_1,x_2,x_3} &amp; x_1^{0.5}x_2^{0.5}+cx_3 \\ \text{s.t.} &amp;x_1+2x_2+px_3\leq w \\ \text{and }&amp; x_1\geq 0, x_2\geq 0, x_3\geq0\end{eqnarray*}</span>
where <span class="math-container">$c&gt;0, p&gt;0, w&gt;0$</span> are given.
It can be re-written as:
<span class="math-container">\begin{eqnarray*} \max_{0\leq x_3\leq w/p} \ \ \max_{x_1\geq 0,x_2\geq 0} &amp; x_1^{0.5}x_2^{0.5}+cx_3 \\ \text{s.t.} &amp; \ x_1+2x_2\leq w-px_3 \end{eqnarray*}</span>
We can solve the problem in two steps. When we solve this:
<span class="math-container">\begin{eqnarray*} \max_{x_1\geq 0,x_2\geq 0} &amp; x_1^{0.5}x_2^{0.5}+cx_3 \\ \text{s.t.} &amp; \ x_1+2x_2\leq w-px_3 \end{eqnarray*}</span>
we get:</p>
<p><span class="math-container">$x_1=\dfrac{w-px_3}{2}$</span> and <span class="math-container">$x_2=\dfrac{w-px_3}{4}$</span>. Therefore, we can write the second step of the problem as:
<span class="math-container">\begin{eqnarray*} \max_{0\leq x_3\leq w/p} &amp; \dfrac{w-px_3+2\sqrt{2}cx_3}{2\sqrt{2}}\end{eqnarray*}</span></p>
<p>and the solution satisfy:
<span class="math-container">\begin{eqnarray*} x_3\in \begin{cases} \left\{\dfrac{w}{p}\right\} &amp; \text{if } 2\sqrt{2}c &gt; p \\ \left[0, \dfrac{w}{p}\right] &amp; \text{if } 2\sqrt{2}c = p \\ \left\{0\right\} &amp; \text{if } 2\sqrt{2}c &lt; p \end{cases}\end{eqnarray*}</span></p>
<p>Given <span class="math-container">$x_3$</span> chosen as above, the corresponding values of <span class="math-container">$x_1, x_2$</span> are
<span class="math-container">\begin{eqnarray*} (x_1,x_2)= \begin{cases} (0,0) &amp; \text{if } 2\sqrt{2}c &gt; p \\ \left(\dfrac{w-px_3}{2},\dfrac{w-px_3}{4}\right) &amp; \text{if } 2\sqrt{2}c = p \\ \left(\dfrac{w}{2},\dfrac{w}{4}\right) &amp; \text{if } 2\sqrt{2}c &lt; p \end{cases}\end{eqnarray*}</span></p>

CC BY-SA 4.0
2023-08-28T05:26:46.673Z
Thanks! This solution is really helpful. I am just wondering if there is any standard reasoning for solving it this way? Say like when facing a quasi-linear utility or some other weird things?
Paul Huang
CC BY-SA 4.0
2023-08-28T06:35:06.980Z
This is a standard reasoning which I use in my lectures for solving such problems. To learn the details, please visit: https://youtube.com/playlist?list=PLUJGfL_499TJ4xlOvpKeAqo4M5xYkq6wi&feature=shared
Amit
CC BY-SA 4.0
2023-08-27T23:04:35.093Z
2023-08-28T07:05:45.083Z
2023-08-28T07:05:45.083Z
Amit
<p>I am dealing with a quasi-linear utility function. For example
<span class="math-container">$U=(x_1x_2)^{0.5}+cx_3$</span> with constrain <span class="math-container">$w\ge x_1+2x_2+px_3$</span>.By taking c, w and p as constant, I function that by using Lagrange multiplier method the F.O.C equations does not have a solution. I just don't know what's the problem. Thanks!</p>
<p>My attempt:</p>
<p>The budget constraint under this case is
<span class="math-container">\begin{equation}
    x_1+2x_2+p_3x_3\le w
\end{equation}</span>
The corresponding Lagrangian (where <span class="math-container">$ \mu $</span> is the Lagrange multiplier) is
<span class="math-container">\begin{equation}
L=\sqrt{x_1x_2}+c x_3+\mu(w-x_1-2x_2-p_3x_3).
\end{equation}</span>
Taking F.O.Cs, we have
<span class="math-container">\begin{equation}
\left\{
\begin{aligned}
    \frac{1}{2}x_2(x_1x_2)^{-\frac{1}{2}}-\mu=0 \\
\frac{1}{2}x_1(x_1x_2)^{-\frac{1}{2}}-2\mu=0 \\
c-p_3\mu=0 \\
w-x_1-2x_2-p_3x_3=0
\end{aligned}
\right.
\end{equation}</span></p>
<p>However, the above set of equations has no solution...</p>

CC BY-SA 4.0
2023-08-27T17:51:29.777Z
Hi! 1. What do you mean when you say this system of equation has no solution?
Giskard
CC BY-SA 4.0
2023-08-27T17:51:52.560Z
2. Have you considered adding non-negativity constraints?
Giskard
CC BY-SA 4.0
2023-08-27T17:52:17.653Z
3. Are you intent on using a Langrangian or would some other method suffice as well?
Giskard
<microeconomics><utility><consumer-theory>
Utility Maximization of a quasi-linear utility function
Paul Huang
# Utility Maximization of a quasi-linear utility function

(Asked by: username_0 on 2023-08-27)

I am dealing with a quasi-linear utility function. For example
$U=(x\_1x\_2)^{0.5}+cx\_3$ with constrain $w\ge x\_1+2x\_2+px\_3$.By taking c, w and p as constant, I function that by using Lagrange multiplier method the F.O.C equations does not have a solution. I just don't know what's the problem. Thanks!

My attempt:

The budget constraint under this case is
\begin{equation}
x\_1+2x\_2+p\_3x\_3\le w
\end{equation}
The corresponding Lagrangian (where $ \mu $ is the Lagrange multiplier) is
\begin{equation}
L=\sqrt{x\_1x\_2}+c x\_3+\mu(w-x\_1-2x\_2-p\_3x\_3).
\end{equation}
Taking F.O.Cs, we have
\begin{equation}
\left\{
\begin{aligned}
\frac{1}{2}x\_2(x\_1x\_2)^{-\frac{1}{2}}-\mu=0 \\
\frac{1}{2}x\_1(x\_1x\_2)^{-\frac{1}{2}}-2\mu=0 \\
c-p\_3\mu=0 \\
w-x\_1-2x\_2-p\_3x\_3=0
\end{aligned}
\right.
\end{equation}

However, the above set of equations has no solution...

**Tags:** microeconomics, utility, consumer-theory
> **Comments:**
> **username_1**: Hi! 1. What do you mean when you say this system of equation has no solution?  
> **username_1**: 2. Have you considered adding non-negativity constraints?  
> **username_1**: 3. Are you intent on using a Langrangian or would some other method suffice as well?  

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## 1 answers

### Answer by username_2 on 2023-08-27. Score: 2

This is the problem we want to solve:
\begin{eqnarray\*} \max\_{x\_1,x\_2,x\_3} & x\_1^{0.5}x\_2^{0.5}+cx\_3 \\ \text{s.t.} &x\_1+2x\_2+px\_3\leq w \\ \text{and }& x\_1\geq 0, x\_2\geq 0, x\_3\geq0\end{eqnarray\*}
where $c>0, p>0, w>0$ are given.
It can be re-written as:
\begin{eqnarray\*} \max\_{0\leq x\_3\leq w/p} \ \ \max\_{x\_1\geq 0,x\_2\geq 0} & x\_1^{0.5}x\_2^{0.5}+cx\_3 \\ \text{s.t.} & \ x\_1+2x\_2\leq w-px\_3 \end{eqnarray\*}
We can solve the problem in two steps. When we solve this:
\begin{eqnarray\*} \max\_{x\_1\geq 0,x\_2\geq 0} & x\_1^{0.5}x\_2^{0.5}+cx\_3 \\ \text{s.t.} & \ x\_1+2x\_2\leq w-px\_3 \end{eqnarray\*}
we get:

$x\_1=\dfrac{w-px\_3}{2}$ and $x\_2=\dfrac{w-px\_3}{4}$. Therefore, we can write the second step of the problem as:
\begin{eqnarray\*} \max\_{0\leq x\_3\leq w/p} & \dfrac{w-px\_3+2\sqrt{2}cx\_3}{2\sqrt{2}}\end{eqnarray\*}

and the solution satisfy:
\begin{eqnarray\*} x\_3\in \begin{cases} \left\{\dfrac{w}{p}\right\} & \text{if } 2\sqrt{2}c > p \\ \left[0, \dfrac{w}{p}\right] & \text{if } 2\sqrt{2}c = p \\ \left\{0\right\} & \text{if } 2\sqrt{2}c < p \end{cases}\end{eqnarray\*}

Given $x\_3$ chosen as above, the corresponding values of $x\_1, x\_2$ are
\begin{eqnarray\*} (x\_1,x\_2)= \begin{cases} (0,0) & \text{if } 2\sqrt{2}c > p \\ \left(\dfrac{w-px\_3}{2},\dfrac{w-px\_3}{4}\right) & \text{if } 2\sqrt{2}c = p \\ \left(\dfrac{w}{2},\dfrac{w}{4}\right) & \text{if } 2\sqrt{2}c < p \end{cases}\end{eqnarray\*}

> **Comments:**
> **username_0**: Thanks! This solution is really helpful. I am just wondering if there is any standard reasoning for solving it this way? Say like when facing a quasi-linear utility or some other weird things?  
> **username_2**: This is a standard reasoning which I use in my lectures for solving such problems. To learn the details, please visit: https://youtube.com/playlist?list=PLUJGfL_499TJ4xlOvpKeAqo4M5xYkq6wi&feature=shared  

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