How to derive the demand function of $U(x_1,x_2)=x_1-x_2$

Question

I'm trying to find the uncompensated demand function. How would you do that, when I have the following utility function $u(x_1,x_2)=x_1-x_2$, and the budget constraint $p_1\cdot x_1+p_2\cdot x_2=p_1\cdot e_1+p_2\cdot e_2$.

I've found the following.

$\max\ u(x_1,x_2)=x_1-x_2 \ s.t. \ p_1(x_1-e_1)+p_2(x_2-e_2)=0$

With use of substitutions I must maximize the following expression

$\max\ u(x_1,x_2)=x_1-\frac{p_1(e_1-x_1)+p_2e_2}{p_2}$

I find the f.o.c.

$\frac{\partial u}{\partial x_1}=1+\frac{p_1}{p_2}=0$

Answer 1

Some economics problems are deisgned to help us practice the basic tools (e.g., optimisation) of the trade. Others are designed to force us to think about the mechanics of a problem to help us understand the underlying economics better.

This problem falls into the latter category, so I recommend you give up on trying to usual calculus (which is premised on finding some sort of indifference curve tangent) and play around with the problem to figure out why this approach isn't working.

---

A consumer with $u=x_1-x_2$ is indifferent between the bundles $(4,3)$ and $(3,2)$. We can plot these and other indifference budles to draw an indifference curve:

Likewise, the consumer is indifferent between $(4,4)$ and $(5,5)$ and we can, in this fashion, get an indifference map:

From inspection of the utility function we know that the consumer is better off with more $x_1$ and less $x_2$, so utility increases in the direction indicated.

Now we can pick an endowment vector (e.g,. $(2,4)$) and a relative price (e.g. $-p_1/p_2=-1$) and plot the price line:

The consumer can trade from their endowment to any point along this line. By inspection, you should be able to figure out what the demand vector $\mathbf{x}=(x_1,x_2)$ will be.

---

Addendum: once you figure out what is going on here, your condition $$\frac{\partial u}{\partial x_1}=1+p_1/p_2$$ should make a bit more sense to you.

Answer 2

If you are studying micro and grasp the notion of Marginal Utilities, you can also notice that for the given utility function $MU_{x_2}=\frac{\partial u}{\partial x_2}<0$ i.e., every additional unit of good 2 reduces your utility. So, it doesn't make sense to consume positive amount of good 2 in optimum.

$$\begin{aligned} \underset{x_1\geq0, x_2 \geq 0}\max & u=x_1-x_2 & \\ \textrm{s.t.} \quad & p_1x_1+p_2x_2=p_1e_1+p_2e_2 \end{aligned}$$

upon substitution $\underset{0 \leq x_1\leq \frac{p_1e_1+p_2e_2}{p_1} }\max u=\frac{x_1(p_1+p_2)}{p_2}-\frac{p_1e_1}{p_2}+e_2$

now as you computed, the objective is increasing in $x_1$ i.e., $\frac{\partial u}{\partial x_1}>0$

So, we set $x_1$ to the maximum value it can take to solve the UMP.

we get, $\boxed{(x_1^*,x_2^*)=(\frac{p_1e_1+p_2e_2}{p_1},0)}$ is your required demand function.

Answer 3

If $p_1, p_2 >0$, you should consume no $x_2$, because it enters your objective negatively. So you should consume $x_1^* = (p_1 e_1 + p_2 e_2)/p_1$ and $x_2^* = 0$.

If the prices are not necessarily positive, you might be able to get infinite utility. For example, if $p_2 = -2$ and $p_1 = 1$, adding 1 unit of $x_2$ relaxes your budget constraint by 2, so you can buy 2 units of $x_1$, and get a net utility increase of 1. You can therefore get infinite utility if $|p_2| > p_1$. There are many cases like this.

Do not use a Lagrangian. The first-order conditions for a Lagrangian are only valid in special cases, the same way that FONCs are not always valid for one-dimensional functions (like $x^3$ on $[0,1]$ or $\mathbb{R}$, for example).