1. How do I show that profit maximization implies cost minimization (in pure competition)?

    Suppose we only consider inputs $l,k$ whose prices are $w,r$ and output price $p$. Profit is $\pi = pf(k,l) - wl - rk$ where $f$ is the production function. Let's assume further that $q^* = f(k^*, l^*)$ maximizes profit.

    To show that this minimizes cost, we need to show that $(k^*, l^*) = \text{arg}\min (wl+rk)$ subject to $f(k,l) = q^*$.

    Is this all we have to show? It seems too trivial, so I am confused.

    But in case that's what we are to show: Assume on the contrary that $(k', l') = \text{arg}\min (wl+rk)$ subject to $f(k,l) = q^*$.

    Then $wl' + rk' < wl^* + rk^* \implies \pi(l',k',q^*) = pq^* - (wl' + rk') > pq^* - (wl' + rk') = \pi(l^*, k^*, q^*)$ contradicting the fact that $(k^*, l^*)$ maximizes profit in the long run.

    Is the proof correct? The fixing of $q^{*}$ is what I am concerned about.

  2. Is it true for monopoly? Does profit maximization imply cost minimization in monopoly?

  • 1
    $\begingroup$ The argument doesn't use the assumption of perfect competition in the output market, so it holds for any market structure. Just replace $p$ by $p(q^*)$. $\endgroup$
    – VARulle
    Nov 2 at 23:45
  • $\begingroup$ @VARulle Yes! (1) My main doubt is if the assumption of $q^{*} = f(k,l)$ is correct. Varian's original solution seems a bit different. Can you tell me why he used $\geq$ instead of $=$? (2) Does "profit maximization imply cost minimization" here refer to "cost minimization at the same ${q^{*}}$ at which profit is maximized"? Thank you! $\endgroup$
    – Rick_Morty
    Nov 3 at 5:10
  • $\begingroup$ Imagine all input prices are equal to $1$ and you have a discrete production technology such that you can produce $6$ widgets by using input vectors $(6, 6)$ or $(7, 2)$, but you can produce $5$ widgets only by using input vector $(5, 5)$. Is it cost-minimizing to use $(5, 5)$ to produce $q^*=5$? With your definition, trivially yes, since it's the only way to do it. With Varian's definition, no. The definitions are equivalent under free disposal (where you can "produce" $5$ widgets also by producing $6$ and throwing away $1$), which is usually assumed, but need not be. $\endgroup$
    – VARulle
    Nov 3 at 8:56
  • $\begingroup$ @VARulle I don't understand that. If $x^{*}$ does not minimize cost of $f(x^{*})$, it means that some other $x$ minimizes the cost of $f(x^{*})$. Written mathematically, $\exists x^{**} \neq x^{*} : x^{**} = \text{arg}\min \text{Cost}(f(x^*))$. Isn't it so? I don't get the use of $\geq$. Can you write it in words instead of an example for me please? I hope I will get it that way. $\endgroup$
    – Rick_Morty
    Nov 3 at 9:46
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    $\begingroup$ It's just two slightly different definitions of what exactly it means to "minimize cost for the output $f(x^*)$". You say it means producing exactly $f(x^*)$ at minimal cost, Varian says it means producing at least $f(x^*)$ at minimal costs. However, it doesn't really matter, since under both definitions, profit maximization implies cost minimization. $\endgroup$
    – VARulle
    Nov 3 at 10:38

1 Answer 1


Is the current production level reached at minimum cost? If not, reach it by lower cost. Revenue is unchanged, hence profit increases, thus it was not maximized.

  • $\begingroup$ Yes, so the proof I wrote is correct? I actually saw Varian's solution manual saying, in the contradiction part: $\exists \ x^{**} : f(x^{**}) \geq f(x^*)$ and $w \cdot x^{*} > w \cdot x^{**}$. I got confused with the $f(x^{**}) \geq f(x^*)$ part as I thought $q^{*} = f(x^*)$ is fixed. $\endgroup$
    – Rick_Morty
    Oct 31 at 17:11
  • $\begingroup$ I am unwilling to comment beyond my proof. $\endgroup$
    – Giskard
    Oct 31 at 21:13

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