When we have the function $U(x_1,x_2) = min\{x_1,3x_2\}$

S.t. $p_1x_1 + p_2x_2 = m$

What's the economic, and mathematical intuition for assuming this constraint is binding, i.e. not having to investigate the case $Px<m$

  1. I was trying to say something like the the utility function is non-decreasing in $x_1$ and $x_2$ but i don't this is strong enough to assume a binding constraint.
  • I understand the function is strictly increasing in the ratio given by Our optimal factor mix, which is along the ray $y = \frac{1}{3}x$. But i don't know how i use this to make my point?
  1. If we were to investigate this case $Px<m$, how would we do it? Where would it go wrong?

  2. What's a good economic statement i can use to also justify ignoring the trivial case of binding non-negativity constraints i.e. $x_1 =0$ or $x_2=0$ apart from just stating that obviously $u = 0$?

  • $\begingroup$ The utility function $U(x_1,x_2)=\min(x_1,3x_2)$ represents preferences that are weakly monotonic this means that the consumer prefers those bundles that contain more of both goods. Consequently, in order to maximize his utility it makes sense for such a consumer to consume the bundle which exhausts all your income. You can also try plotting the budget set and the level curves to see that the optimum lies on the budget line $\endgroup$
    – mynameparv
    May 24, 2023 at 11:51

2 Answers 2


Suppose there is an optimal bundle $x^\star = ({x_1}^\star,{x_2}^\star)$ such that $p \cdot x^\star < m$.

Since $p \cdot x^\star < m$, there exists a $t > 1$ such that $t (p \cdot x^\star) = m$.

More specifically, $t = \frac{m}{p \cdot x^\star}$.

Therefore, $p \cdot (tx^\star) = m$.

Plugging into the utility function,

$U(tx^\star) = U(t {x_1}^\star, t {x_2}^\star) = \min\{t {x_1}^\star, 3 t {x_2}^\star \} $

Since $t > 1 > 0$,

$U(tx^\star) = t \min \{{x_1}^\star, 3 {x_2}^\star\} = t U(x^\star)$

Since $t>1$ and $U(x^\star) > 0$,

$U(tx^\star) = t U(x^\star) > U(x^\star)$

Since $tx^\star$ is a strictly preferred bundle to $x^\star$, this contradicts that $x^\star$ is an optimal bundle.

Since $x^\star$ was taken as an arbitrary optimal bundle such that $p \cdot x^\star < m$, we can conclude that no such optimal bundle can exist.

Therefore, an optimal bundle for this utility must satisfy $p \cdot x^\star = m$.

Intuitively, you can scale up the bundle proportionately until you spend all your budget, yielding a strictly better bundle.

  • $\begingroup$ Wow, love that you just thought this up! Im starting to wonder if there are questions you don’t know the answer to lol $\endgroup$
    – CormJack
    May 23, 2023 at 23:30
  • $\begingroup$ Also you use the homogeneity of the min function. Just checking that we could still show homogeneity if $0<t<1$. But the reason we need $t > 1$ is to provide our deserved contradiction! $\endgroup$
    – CormJack
    May 23, 2023 at 23:35
  • $\begingroup$ I actually love your answer, it’s so clean. It feels like the kind of proof you see in a book, where afterwards it looks so obvious. But before hand it just doesn’t occur to you at all hahaha $\endgroup$
    – CormJack
    May 23, 2023 at 23:37


Suppose there is an optimal bundle $x^\star$ such that $p \cdot x^\star < m$. By continuity there is a neighborhood $N$ of $x^\star$ such that $p\cdot x<m$ and $u(x)\le u(x^\star)$ for all $x\in N$. This contradicts local non-satiation. Thus, no bundle in the interior of the budget set can be optimal.

Mathematical intuition:

Monotone preferences like the given one are locally non-satiated, hence for every bundle $x^\star$ there are always arbitrarily close better bundles. If $x^\star$ is in the interior of the budget set, then these better bundles are also in the budget set, so $x^\star$ cannot be optimal.

Economic intuition:

If you buy the optimal bundle $x^\star$ and $p \cdot x^\star < m$, then you have some money left over. You could have used this money to buy stuff and thereby further increase your utility. But then $x^\star$ cannot be optimal.


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