2
$\begingroup$

Let an agent with utility function $U(x_1,x_2,x_3)=min\{ x_1, x_2 \} +x_3$. We want to maximize $U$ subject to $p_1 x_1+p_2 x_2 +p_3 x_3= I$

First intuitively $x_1^*=x_2^*$ substituting en the budget constraint I have $$x_1^*=x_2^*=\frac{I-p_3 x_3}{p_1+p_2}$$ I don't know how to solve for $x_3$. Then i thought in Kuhn-Tucker but I can't see how to solve it.

$\endgroup$

2 Answers 2

1
$\begingroup$

I will give a hint. To increase/decrease utility by 1 unit, you can do two things. Increase $x_1$ and $x_2$ by $\frac{1}{2}$ units and this would cost you $\frac{p_1+p_2}{2}$. The second option is changing $x_3$ by one unit that would cost $p_3$. One other option is of course using a linear combination of the above changes. If you remember from well behaved utility functions, you would like to buy things for which utility per price is higher. Now you would go for a similar reasoning here and would try to buy extra utility at a cheapest price.

$\endgroup$
1
  • $\begingroup$ I think i get it. Thank you, if something happens I shall ask again. $\endgroup$ Sep 30, 2015 at 0:17
1
$\begingroup$

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:

enter image description here

  • $p_1 + p_2 > p_3 \rightarrow$ 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:

enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.