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enter image description here

Those two are generic production functions that we usually see, and I check the definition of quasiconvex and quasiconcave on wikipedia, and it seems those two graphs satisfy both definitions, and then so-called quasilinear. Then, I ponder why the production function we would generally assume to be quasiconcave only. Or I might misunderstand the definition or carelessly draw the wrong graph?

Appreciated any comment and help

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  • $\begingroup$ Is it a function with a single variable, $F(K)$? $\endgroup$
    – Giskard
    Aug 16 at 15:56
  • $\begingroup$ Is it also monotonically increasing? $\endgroup$
    – Giskard
    Aug 16 at 15:56
  • $\begingroup$ @Giskard Thank you for your reply. Both Yes. But I have no idea why these questions are related to quasiconcavity. What happens if we suppose they are two variables function, let's considered it as a 3D picture. are they still quasi-convex? $\endgroup$
    – LJNG
    Aug 16 at 16:03

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Think of "quasiconcave" as single-peak, and "quasiconvex" as single-trough. A function that has a single-peak has a unique maximum/supremum (both local and global), and a function that has a single-trough has a unique minimum/infimum (again both local and global). Clearly the two do not contradict one another. The production function is assumed to be quasiconcave and not necessarily quasiconvex because we're more interested how it peaks, i.e. reaches a maximum as in revenue maximization, than how it reaches a minimum.

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  • $\begingroup$ Thank you for your reply. Is my two graphs also quasilinear, they seems also fit both difinition $f(tx+(1-t)y)\leq max\{f(x),f(y)\}$ and $f(tx+(1-t)y)\geq min\{f(x),f(y)\}$ $\endgroup$
    – LJNG
    Aug 17 at 1:21
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    $\begingroup$ I disagree with thinking of quasiconcave as single-peaked. A monotonically increasing function as in the question admits no local maximum and is therefore not single-peaked, but any such function is quasiconcave. (Actually the concept of quasiconcavity is not very useful for increasing functions on a single dimension.) $\endgroup$
    – VARulle
    Aug 17 at 11:22
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    $\begingroup$ @Giskard: That's a good question and I don't have a ready answer at the moment. Please do make a separate post on it. $\endgroup$
    – Herr K.
    Aug 17 at 21:57
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    $\begingroup$ Done! $\endgroup$
    – Giskard
    Aug 18 at 8:19
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    $\begingroup$ Wow, I just figured out how to write comments shorter than 15 chars, very proud of myself. $\endgroup$
    – Giskard
    Aug 18 at 8:20

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