# Does profit maximization imply cost minimization in both pure competition and monopoly?

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?

• 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 $p(q^*)$. Commented Nov 2, 2022 at 23:45
• @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! Commented Nov 3, 2022 at 5:10
• 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. Commented Nov 3, 2022 at 8:56
• @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. Commented Nov 3, 2022 at 9:46
• 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. Commented Nov 3, 2022 at 10:38

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.

• 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. Commented Oct 31, 2022 at 17:11
• I am unwilling to comment beyond my proof. 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!!