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

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?

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 $$B$$?

• If I understand you correctly, you buy at bar $A$ for $p1+p2$, and at bar $B$ for $p3$. Oct 9, 2019 at 18:55
• That is correct, but what I meant with the last sentence was: At what values of $p_1$, $p_2$, $p_3$ is it cheaper to buy my coffee with sugar at $A$? Oct 9, 2019 at 19:36
• I edited the answer to make it more clear, try to relate the little story above to the problem you have to solve Oct 9, 2019 at 19:39
• Thank you. In that case it means that you will buy at $A$ if $p1 + p2$ < $p3$. If $p1 + p2$ > $p3$, you will buy at $B$. If they are equal, you are indifferent. Oct 9, 2019 at 20:09
• Exactly! And if you now combine this with the fact that your budget is equal to $w$ you have your demand function. Oct 9, 2019 at 21:09

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

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

The demands of these then proceed how you would for any case of perfect substitutes. However you must list them for each case.

Hope this helps

We solve the problem

$$$$\max U(x_1,x_2,x_3) = \min\{x_1,x_2\} + x_3$$$$

subject to

$$$$x_1 p_1 + x_2 p_2 + x_3 p_3 = I$$$$

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

$$$$x_1 (p_1 + p_2) + x_3 p_3 = I$$$$

Solving for $$x_1$$ we get

$$$$x_1 = \frac{I - x_3 p_3}{p_1 + p_2}$$$$

Since $$x_1 = x_2$$,

$$$$x_2 = \frac{I - x_3 p_3}{p_1 + p_2}$$$$

Therefore, the min term becomes

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

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

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

Separating the fraction and rearranging we get

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

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:

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.