Derive demand function $x(p,w)$ from utility function $u(x) = \min\{x_1, x_2\} + x_3$

Question

I know how to solve the two-good case with $u(x) = \min\{x_1, x_2\}$, but the addition of $x_3$ confuses me.

Problem

Derive the demand function $x(p,w)$ from $u(x) = \min\{x_1, x_2\} + x_3$.

What I did so far

We assume that in optimum $x_1 = x_2$.

Set up the budget constraint $p_1x_1 + p_2x_2 + p_3x_3 = w$.

Rewrite budget constraint as $(p_1+p_2)x_1 + p_3x_3 = w$ or $(p_1+p_2)x_2 + p_3x_3 = w$.

We can write $x_1^*=x_2^* = \frac{w-p_3x_3}{p_1+p_2}$ and $x_3^* = \frac{w-(p_1+p_2)x_1^*}{p_3}$.

Confusion

How to proceed? Can I still use a Lagrangian to solve this?

Answer 1

Hint: Imagine that there are two coffee bars, $A$ and $B$. There is only one type of coffee in the world. My preferences are such that I always want 1 unit of sugar with 1 unit of coffee; if I consume units of coffee and sugar in the ratio $1:1$, additional units of only one of the two don't give me extra utility.

At coffee bar $A$ they sell coffee and sugar separately. That is: there is one counter for coffee, where coffee is sold for a price of $p_1$ per unit. There is another counter for sugar, where I can buy one unit of sugar for a price of $p_2$.

At coffee bar $B$ however, they only sell one unit of coffee with one unit of sugar in a bundle, the price of which is equal to $p_3$. They do not sell anything separately

At what prices do I prefer to buy my coffee with sugar at $A$, at what prices do I prefer to buy my coffee with sugar at $B$?

Answer 2

We solve the problem

\begin{equation} \max U(x_1,x_2,x_3) = \min\{x_1,x_2\} + x_3 \end{equation}

subject to

\begin{equation} x_1 p_1 + x_2 p_2 + x_3 p_3 = I \end{equation}

From the min term we get that $x_1 = x_2$. Therefore the budget constraint becomes

\begin{equation} x_1 (p_1 + p_2) + x_3 p_3 = I \end{equation}

Solving for $x_1$ we get

\begin{equation} x_1 = \frac{I - x_3 p_3}{p_1 + p_2} \end{equation}

Since $x_1 = x_2$,

\begin{equation} x_2 = \frac{I - x_3 p_3}{p_1 + p_2} \end{equation}

Therefore, the min term becomes

\begin{equation} \min\{x_1,x_2\} = \frac{I - x_3 p_3}{p_1 + p_2} \end{equation}

Substituting this expression for the min term into the utility function we get

\begin{equation} W(x_3) := U(x_1(x_3),x_2(x_3),x_3) = \frac{I - x_3 p_3}{p_1 + p_2} + x_3 \end{equation}

Separating the fraction and rearranging we get

\begin{equation} W(x_3) = (1 - \frac{p_3}{p_1 + p_2}) x_3 + \frac{I}{p_1 + p_2} \end{equation}

Note that this is a straight line, with slope $1 - \frac{p_3}{p_1 + p_2}$.

Notice $1 - \frac{p_3}{p_1 + p_2} > 0 \iff 1 > \frac{p_3}{p_1 + p_2} \iff p_1 + p_2 > p_3$.

From here we get 3 cases:

• $p_1 + p_2 > p_3 \rightarrow$ spend everything on $x_3 \rightarrow x_1 = 0, x_2 = 0, x_3 = \frac{I}{p_3}$.
• $p_1 + p_2 < p_3 \rightarrow$ don't consume $x_3 \rightarrow x_1 = \frac{I}{p_1 + p_2}, x_2 = \frac{I}{p_1 + p_2}, x_3 = 0$
• $p_1 + p_2 = p_3 \rightarrow x_3$ value indifferent $\rightarrow x_1 = \frac{I - x_3 p_3}{p_1 + p_2}, x_2 = \frac{I - x_3 p_3}{p_1 + p_2}, 0 \leq x_3 \leq \frac{I}{p_3}$, i.e. the optimal bundles form a line in 3-D space that looks like this:

Answer 3

for this problem you must consider two possible branches of the utility function:

$$u(\text{x})=x_1+x_3\ \ \text{if} \ \ x_1<x_2$$ $$u(\text{x})=x_2+x_3\ \ \text{if} \ \ x_1>x_2$$

The demand equations you solve from each one of these provide guidance on what the solution would be to your problem. However you must list them for each case.

Hope this helps

Answer 4

Since you already know how to solve for the two goods case with $u(x)=\min(x_1,x_2)$ you can solve the optimization problem in a two-stage method without using Lagrangian.

UMP: $$\begin{aligned} \max_{x_1,x_2,x_3\geq 0} \quad & \min(x_1,x_2)+x_3\\ \textrm{s.t.} \quad & p_1x_1+p_2x_2+p_3x_3\leq M\end{aligned}$$

Stage one: Solving the problem with respect to $x_1$ and $x_2$ holding $x_3$ constant $$\max_{x_3}\begin{pmatrix}{\begin{aligned}\max_{x_1,x_2} \quad & \min(x_1,x_2)+x_3\\ \textrm{s.t. } \quad & p_1x_1+p_2x_2\leq M-p_3x_3\end{aligned}}\end{pmatrix}$$

The problem inside the parentheses yields: $x_1(p,x_3,M)=x_2(p,x_3,M)=\frac{M-p_3x_3}{p_1+p_2}$ Notice that the solution to the problem inside the brackets is similar to a demand function for $u=\min(x_1,x_2)$ with the difference being that we adjusted income in the demand function for the expenditure that might occur on $x_3$

Stage Two: Solving the entire problem in terms of $x_3$ after obtaining demands for $x_1,x_2$ as a function of $x_3$. Substituting the values of $x_1,x_2$ that we derived above in the problem inside the parentheses gives us the following optimization problem in $x_3$: $$\begin{aligned} \max_{0\leq x_3\leq \frac{M}{p_3}} \quad & x_3+\frac{M-p_3x_3}{p_1+p_2} \\ \max_{0\leq x_3\leq \frac{M}{p_3}} \quad & \frac{(p_1+p_2-p_3)x_3}{p_1+p_2}+\frac{M}{p_1+p_2} \end{aligned}$$

from the above problem, we get the demand function for $x_3$ and consequently the demand functions for $x_1, x_2$ by substituting the demand for $x_3$ in our solution to the Stage One problem.

$\boxed{(x_1,x_2,x_3)^d(p,M)=\begin{cases}(\frac{M}{p_3},0,0) & \text{if } p_1+p_2>p_3\\ (\frac{M}{p_1+p_2},\frac{M}{p_1+p_2},0) & \text{if } p_1+p_2\leq p_3\end{cases}}$

Intuitively it is like a solution to a perfect substitute utility function, but one of the goods is a composite good consisting of two goods which are complementary.