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.

  • $\begingroup$ Why did you delete most of the question? Without the question, the answer by tdm is not as helpful it would otherwise be. $\endgroup$
    – smcc
    Sep 4, 2023 at 14:44
  • 1
    $\begingroup$ @smcc sorry about that. I accidentally deleted the question. Now it's restored. $\endgroup$ Sep 4, 2023 at 14:54

1 Answer 1


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.

  • $\begingroup$ what if $u_{21}=u_{12}>0$? $\endgroup$ Sep 4, 2023 at 19:43
  • $\begingroup$ 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. $\endgroup$
    – tdm
    Sep 5, 2023 at 5:16

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