# Quasi-concavity of profit function

I'm doing a simple exercise from my textbook: suppose the revenue function is $$R(p)=p^{1-\epsilon}$$ with $$\epsilon > 0$$. Suppose the cost function is convex. Show that the profit function is quasi-concave if $$\epsilon > 1$$.

My attempt: Let the demand function then be $$q(p)=p^{-\epsilon}$$ and the cost function some $$c(q(p))$$.

The first-order condition is $$(1-\epsilon)p^{-\epsilon}-c'(q(p))q'(p)$$ and the second order condition $$-\epsilon(1-\epsilon)p^{-\epsilon-1}-c''(q(p))q'(p)^2-c'(q(p))q''(p)$$.

The second term $$-c''(q(p))q'(p)^2$$ and third terms $$-c'(q(p))q''(p)$$ are negative under convexity of the cost function. Hence the second order condition is unambiguously negative if the first term is negative. But that happens for $$\epsilon<1$$, the opposite as conjectured.

• Perhaps edit your calculations into your question, that way people can see what goes wrong. – Giskard Apr 11 '20 at 11:31
• Done, thank you. – econ86 Apr 11 '20 at 12:13
• Yes, the second derivative may be negative, but so what? E.g., for $\epsilon = 1/2$ and $C(y) = 0$ you get $\Pi(p) = p^{1/2}$. Is this quasi-concave? – Giskard Apr 11 '20 at 12:39
• Well, I was trying to show that the objective is concave in prices, hence quasi-concave. How would you proceed instead to get that $\epsilon > 1$ result? – econ86 Apr 11 '20 at 12:48
• I know the definition in terms of the $max$ operator and that of contour sets, but honestly don't know how to apply them there. A hint or an answer would be appreciated. – econ86 Apr 11 '20 at 14:26

Let the demand function be $$D(p) = p^{-\epsilon}$$ and the cost function some $$C(D(p))$$ so that $$R(p) = D(p)p = p^{1-\epsilon}$$, right?

Then, I'd rewrite your first and second order conditions - as you only provided the expressions but it would be great to write them in a form of equations (or inequalities). So that:

First order condition: $$R'(p) - C'(D(p))D'(p) =0$$

And the second order: $$R''(p) - C''\left(D(p)\right)\cdot D'(p)^2 - C'(D(p))D''(p) < 0$$

This is the moment when we kind of split our approaches. I also do agree that the second term in second order condition is negative. However, it is not necessarily $$R''(p)$$ that is negative - rather the whole relation $$R''(p) - C'(D(p))D''(p)<0 \;(*)$$ Now, we want the second order condition to hold for every $$p$$ that satisfies the first order condition. Thus, I would substitute the $$C'(D(p))$$ from the first condition to the $$(*)$$ and hence, I end up with: $$R''(p) - \frac{R'(p)D''(p)}{D'(p)} <0 \;(**)$$

Now it's time for the brute force. Calculate all derivatives in $$(**)$$.

$$-\epsilon(1-\epsilon)p^{-1-\epsilon} - \frac{(1-\epsilon)p^{-\epsilon} \cdot -\epsilon(-1-\epsilon)p^{-2-\epsilon}}{-\epsilon p^{-1-\epsilon}} <0 \implies \epsilon > 1$$ for both, $$p$$ and $$\epsilon$$ positive.

P.S. If you think about it, letting $$\epsilon<1$$ would imply that $$R'(p)>0$$ and thus, the first order condition would have no solutions. Why?

P.P.S. I believe a similar exercise is in Industrial Organization: Markets and Strategies by Belleflamme and Peitz in their chapter about "monopoly pricing strategies" but with the inverse demands $$P(q)$$.

• Thank you very much! This one is from The theory of industrial organization by Tirole. The chapters on monopoly pricing are quite similar to one another. I had seen the one you mention too when looking for answers! – econ86 Apr 12 '20 at 15:04