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
    Commented Nov 2, 2022 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
    Commented Nov 3, 2022 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
    Commented Nov 3, 2022 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
    Commented Nov 3, 2022 at 9:46
  • 1
    $\begingroup$ 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
    Commented Nov 3, 2022 at 10:38

2 Answers 2


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
    Commented Oct 31, 2022 at 17:11
  • $\begingroup$ I am unwilling to comment beyond my proof. $\endgroup$
    – Giskard
    Commented Oct 31, 2022 at 21:13

Hey hope this helps you out. The maximization problem is:

$$\max_{x_1,x_2,y} py - w_1x_1 - w_2x_2$$

$s.t. y=f(x_1,x_2)$

In order to maximize, we could first choose to maximize with respect to $x_1$ then to $x_2$ and later to $y$. So we are going to use this in our advantage. First, we are going to maximize with respect to $x_1,x_2$ and later on with respect to $y$.

First Step:

$$\max_{x_1,x_2} py - w_1x_1 - w_2x_2$$

$s.t. y=f(x_1,x_2)$

Because we are not choosing $y$, its like having a constant value. Then $py$ is constant we could take it off for a moment, and our problem would be:

$$\max_{x_1,x_2} - w_1x_1 - w_2x_2= -(w_1x_1 + w_2x_2)$$

$s.t. y=f(x_1,x_2)$

Because we are maximizing $-(w_1x_1 + w_2x_2)$ we could write it like minimizing $w_1x_1 + w_2x_2$. We have arrived to our minimizing problem. And we would get $C^*(w_1,w_2,y)$.

Second Step:

$$\max_{y} py - C^*(w_1,w_2,y)$$

$$\max_{y} py- C^*(w_1,w_2,y)\equiv \max_{x_1,x_2,y} pf(x_1,x_2) - w_1x_1 - w_2x_2$$

And these problems are equivalent. We have just altered the order of maximizing. That's why the maximization of profit problem implies minimizing the cost!!


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.