<p>Given that <span class="math-container">$u_{1,2} = 0$</span>, the utility function is additively separable. So we have the problem:
<span class="math-container">$$
\begin{align*}
\max_{x_1, x_2} u(x_1) + v(x_2) \text{ s.t. } &amp;p_1 x_1 + p_2 x_2 = m \ell\\
&amp; k t_1 x_1 + k t_2 x_2 = T-\ell
\end{align*}
$$</span>
The first constraint is the budget constraint. The second constraint is the time constraint.</p>
<p>Substiting the second constaint into the first (eliminating <span class="math-container">$\ell$</span>) gives:
<span class="math-container">$$
(p_1 + m k t_1) x_1 + (p_2 + m k t_2) x_2 = mT
$$</span></p>
<p>Now, the first order conditions for the maximisation problem give:
<span class="math-container">$$
\begin{align*}
&amp;u'(x_1) = \lambda (p_1 + m kt_1)\\
&amp;v'(x_2) = \lambda (p_2 + m k t_2)\\
&amp;(p_1 + mkt_1) x_1 + (p_2 + m k t_2) x_2 = m T
\end{align*}
$$</span>
we can totally differentiate the three equations (with variables <span class="math-container">$x_1, x_2, \lambda$</span> and <span class="math-container">$k$</span>).</p>
<p><span class="math-container">$$
\begin{align*}
&amp;u''(x_1) dx_1 - (p_1 + m kt_1) d \lambda = \lambda m t_1 dk\\
&amp;v''(x_2) dx_2 - (p_2 + m kt_2) d \lambda = \lambda m t_2 dk\\
&amp;(p_1 + mkt_1) dx_1 + (p_2 + m k t_2) dx_2 = -m(t_1 x_1 + t_2 x_2) dk.
\end{align*}
$$</span></p>
<p>In matrix form, this gives:
<span class="math-container">$$
\begin{bmatrix}u''(x_1) &amp; 0 &amp; -(p_1 + m kt_1)\\
0 &amp; v''(x_2) &amp; - (p_2 + m k t_2)\\
(p_1 + m k t_1) &amp; (p_2 + m k t_2) &amp; 0 \end{bmatrix} \begin{bmatrix} dx1 \\ dx2 \\ d \lambda\end{bmatrix} = \begin{bmatrix} \lambda m t_1 \\ \lambda m t_2 \\ -m(t_1 x_1 + t_2 x_2) \end{bmatrix} dk
$$</span></p>
<p>We can solve this system:
<span class="math-container">$$
\begin{align*}
\begin{bmatrix} dx_1 \\ dx_2 \\ d \lambda\end{bmatrix} = \frac{A}{v''(x_2)(p_1 + m k t_1)^2 + u''(x_1)(p_2 + m k t_2)^2}  \begin{bmatrix}\lambda m t_1\\ \lambda m t_2 \\-m(t_1 x_1 + t_2 x_2)\end{bmatrix} d k
\end{align*}
$$</span>
where <span class="math-container">$A$</span> is the matrix:
<span class="math-container">$$
A = \begin{bmatrix}(p_2 + m k t_2)^2 &amp; -(p_1 + m k t_1)(p_2 + m k t_2) &amp; v''(x_2)(p_1 + mkt_2)\\
-(p_1 + m k t_1)(p_2 + mkt_2) &amp; (p_1 + m k t_1)^2 &amp; u''(x_1)(p_2 + m k t_2) \\
-v''(x_2)(p_1 + m k t_1) &amp; -u''(x_1)(p_2 + m k t_2) &amp; u''(x_1)v''(x_2)\end{bmatrix}
$$</span></p>
<p>Now, taking the second row, we have:
<span class="math-container">$$
dx_2 = \frac{-(p_1 + m k t_1)(p_2 + m k t_2)\lambda mt_1 + (p_1 + mkt_1)^2 \lambda m t_2 + u''(x_1)(p_2 + m k t_2)(-m(t_1 x_1 + t_2 x_2))}{v''(x_2)(p_1 + m k t_1)^2 + u''(x_1)(p_2 + m k t_2)^2}dk
$$</span></p>
<p>The denominator is negative as <span class="math-container">$u''(x_1), v''(x_2) &lt; 0$</span>. For the numerator, the third term reduces to:
<span class="math-container">$$
-u''(x_1)(p_2 + m k t_2)(t_1 x_1 + t_2 x_2) &gt; 0.
$$</span>
The first two terms give:
<span class="math-container">$$
\begin{align*}
&amp;-(p_1 + m kt_1)(p_2 + m k t_2)\lambda m t_1 + (p_1 + m k t_1)^2 \lambda m t_2,\\
&amp;= (p_1 + m kt_1)\lambda m(-p_2t_1 -  m k t_2t_1 + p_1 t_2 + m k t_1 t_2)\\
&amp;= (p_1 + m k t_1)\lambda m(p_1 t_2 - p_2 t_1)
\end{align*}
$$</span>
The latter is positive if <span class="math-container">$\frac{p_2}{p_1} \le \frac{t_2}{t_1}$</span>. If so, the numerator of <span class="math-container">$\frac{dx_2}{dk}$</span> is positive and the denominator is negative so the overal sign is negative.</p>

CC BY-SA 4.0
2023-09-04T19:43:20.300Z
what if $u_{21}=u_{12}>0$?
Ludwig Gershwin
CC BY-SA 4.0
2023-09-05T05:16:11.933Z
Then the total derivatives change. For example, the first becomes $u_{11} dx_1 + u_{12} dx_2-(p_1 + mkt_1) d \lambda = \lambda m t_1 dk$ and so on. For the further analysis, taking the matrix inverse becomes somewhat more complicated but I think it can be done.
tdm
CC BY-SA 4.0
2023-09-04T06:10:26.457Z
2023-09-04T06:31:12.213Z
2023-09-04T06:31:12.213Z
tdm
<p>Suppose a consumer has a utility function <span class="math-container">$u(x_1,x_2)$</span>, where <span class="math-container">$u_1&gt;0, u_2&gt;0, u_{11}&lt;0,u_{22}&lt;0,u_{12}=u_{21}&gt;0$</span>. The prices are fixed. The consumer's income comes from working and the wage is <span class="math-container">$m$</span> per hour. The consumer also has <span class="math-container">$T$</span> hours in a day to allocate between work and consumption. A unit of good <span class="math-container">$i$</span> takes <span class="math-container">$kt_i$</span> hours to consume.</p>
<p>Suppose that when the consumer is healthy, <span class="math-container">$k=1$</span>; but when the consumer is not healthy, it takes more time to consume. I want to show that if <span class="math-container">$\frac{p_2}{p_1}\leq\frac{t_2}{t_1}$</span>, then the consumption of good 2 falls when the consumer's health is not good; that is, when <span class="math-container">$k$</span> rises above 1.</p>
<p>Seems like I need to do total differentiation in order to solve this problem, but I have no idea where to start. Anyone has any thoughts?</p>
<p>Edited: <span class="math-container">$u_{21}=u_{12}&gt;0$</span>. My bad.</p>

CC BY-SA 4.0
2023-09-04T14:44:33.330Z
Why did you delete most of the question? Without the question, the answer by tdm is not as helpful it would otherwise be.
smcc
CC BY-SA 4.0
2023-09-04T14:54:26.243Z
@smcc sorry about that. I accidentally deleted the question. Now it's restored.
Ludwig Gershwin
<utility><comparative-statics>
How do I use total derivatives of an implicit function to solve this problem?
Ludwig Gershwin
# How do I use total derivatives of an implicit function to solve this problem?

(Asked by: username_0 on 2023-09-03)

Suppose a consumer has a utility function $u(x\_1,x\_2)$, where $u\_1>0, u\_2>0, u\_{11}<0,u\_{22}<0,u\_{12}=u\_{21}>0$. The prices are fixed. The consumer's income comes from working and the wage is $m$ per hour. The consumer also has $T$ hours in a day to allocate between work and consumption. A unit of good $i$ takes $kt\_i$ hours to consume.

Suppose that when the consumer is healthy, $k=1$; but when the consumer is not healthy, it takes more time to consume. I want to show that if $\frac{p\_2}{p\_1}\leq\frac{t\_2}{t\_1}$, then the consumption of good 2 falls when the consumer's health is not good; that is, when $k$ rises above 1.

Seems like I need to do total differentiation in order to solve this problem, but I have no idea where to start. Anyone has any thoughts?

Edited: $u\_{21}=u\_{12}>0$. My bad.

**Tags:** utility, comparative-statics
> **Comments:**
> **username_1**: Why did you delete most of the question? Without the question, the answer by tdm is not as helpful it would otherwise be.  
> **username_0**: @smcc sorry about that. I accidentally deleted the question. Now it's restored.  

---

## 1 answers

### Answer by username_2 on 2023-09-04. Score: 2 | Accepted answer

Given that $u\_{1,2} = 0$, the utility function is additively separable. So we have the problem:
$$
\begin{align\*}
\max\_{x\_1, x\_2} u(x\_1) + v(x\_2) \text{ s.t. } &p\_1 x\_1 + p\_2 x\_2 = m \ell\\
& k t\_1 x\_1 + k t\_2 x\_2 = T-\ell
\end{align\*}
$$
The first constraint is the budget constraint. The second constraint is the time constraint.

Substiting the second constaint into the first (eliminating $\ell$) gives:
$$
(p\_1 + m k t\_1) x\_1 + (p\_2 + m k t\_2) x\_2 = mT
$$

Now, the first order conditions for the maximisation problem give:
$$
\begin{align\*}
&u'(x\_1) = \lambda (p\_1 + m kt\_1)\\
&v'(x\_2) = \lambda (p\_2 + m k t\_2)\\
&(p\_1 + mkt\_1) x\_1 + (p\_2 + m k t\_2) x\_2 = m T
\end{align\*}
$$
we can totally differentiate the three equations (with variables $x\_1, x\_2, \lambda$ and $k$).

$$
\begin{align\*}
&u''(x\_1) dx\_1 - (p\_1 + m kt\_1) d \lambda = \lambda m t\_1 dk\\
&v''(x\_2) dx\_2 - (p\_2 + m kt\_2) d \lambda = \lambda m t\_2 dk\\
&(p\_1 + mkt\_1) dx\_1 + (p\_2 + m k t\_2) dx\_2 = -m(t\_1 x\_1 + t\_2 x\_2) dk.
\end{align\*}
$$

In matrix form, this gives:
$$
\begin{bmatrix}u''(x\_1) & 0 & -(p\_1 + m kt\_1)\\
0 & v''(x\_2) & - (p\_2 + m k t\_2)\\
(p\_1 + m k t\_1) & (p\_2 + m k t\_2) & 0 \end{bmatrix} \begin{bmatrix} dx1 \\ dx2 \\ d \lambda\end{bmatrix} = \begin{bmatrix} \lambda m t\_1 \\ \lambda m t\_2 \\ -m(t\_1 x\_1 + t\_2 x\_2) \end{bmatrix} dk
$$

We can solve this system:
$$
\begin{align\*}
\begin{bmatrix} dx\_1 \\ dx\_2 \\ d \lambda\end{bmatrix} = \frac{A}{v''(x\_2)(p\_1 + m k t\_1)^2 + u''(x\_1)(p\_2 + m k t\_2)^2} \begin{bmatrix}\lambda m t\_1\\ \lambda m t\_2 \\-m(t\_1 x\_1 + t\_2 x\_2)\end{bmatrix} d k
\end{align\*}
$$
where $A$ is the matrix:
$$
A = \begin{bmatrix}(p\_2 + m k t\_2)^2 & -(p\_1 + m k t\_1)(p\_2 + m k t\_2) & v''(x\_2)(p\_1 + mkt\_2)\\
-(p\_1 + m k t\_1)(p\_2 + mkt\_2) & (p\_1 + m k t\_1)^2 & u''(x\_1)(p\_2 + m k t\_2) \\
-v''(x\_2)(p\_1 + m k t\_1) & -u''(x\_1)(p\_2 + m k t\_2) & u''(x\_1)v''(x\_2)\end{bmatrix}
$$

Now, taking the second row, we have:
$$
dx\_2 = \frac{-(p\_1 + m k t\_1)(p\_2 + m k t\_2)\lambda mt\_1 + (p\_1 + mkt\_1)^2 \lambda m t\_2 + u''(x\_1)(p\_2 + m k t\_2)(-m(t\_1 x\_1 + t\_2 x\_2))}{v''(x\_2)(p\_1 + m k t\_1)^2 + u''(x\_1)(p\_2 + m k t\_2)^2}dk
$$

The denominator is negative as $u''(x\_1), v''(x\_2) < 0$. For the numerator, the third term reduces to:
$$
-u''(x\_1)(p\_2 + m k t\_2)(t\_1 x\_1 + t\_2 x\_2) > 0.
$$
The first two terms give:
$$
\begin{align\*}
&-(p\_1 + m kt\_1)(p\_2 + m k t\_2)\lambda m t\_1 + (p\_1 + m k t\_1)^2 \lambda m t\_2,\\
&= (p\_1 + m kt\_1)\lambda m(-p\_2t\_1 - m k t\_2t\_1 + p\_1 t\_2 + m k t\_1 t\_2)\\
&= (p\_1 + m k t\_1)\lambda m(p\_1 t\_2 - p\_2 t\_1)
\end{align\*}
$$
The latter is positive if $\frac{p\_2}{p\_1} \le \frac{t\_2}{t\_1}$. If so, the numerator of $\frac{dx\_2}{dk}$ is positive and the denominator is negative so the overal sign is negative.

> **Comments:**
> **username_0**: what if $u_{21}=u_{12}>0$?  
> **username_2**: Then the total derivatives change. For example, the first becomes $u_{11} dx_1 + u_{12} dx_2-(p_1 + mkt_1) d \lambda = \lambda m t_1 dk$ and so on. For the further analysis, taking the matrix inverse becomes somewhat more complicated but I think it can be done.  

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