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LeoMessi
 2 years ago
Best ResponseYou've already chosen the best response.0\[q = 10K^{2/3} + L^{1/2}\] K = # machines, L = # labor units.

dpflan
 2 years ago
Best ResponseYou've already chosen the best response.3OK, there are (i.) decreasing returns, (ii.) increasing returns, and (iii.) constant returns. Your function is one of K and L so \[q=f(K,L)\]. if we increase K and L by 2, is the quantity produced greater than, less than, or equal to 2 * the production function?

dpflan
 2 years ago
Best ResponseYou've already chosen the best response.3\[2*F(K,L) versus F(2*K, 2*L)\]

LeoMessi
 2 years ago
Best ResponseYou've already chosen the best response.0\[F(2*K, 2*L) = 10*(2K)^{2/3} + (2*L)^{1/2} = 10 * 2^{2/3} * K + 2^{1/2} * L\] and \[2*F(K,L) = 2 *(10 * K^{2/3} + L^{1/2}\]

dpflan
 2 years ago
Best ResponseYou've already chosen the best response.3right, to easily compare the 2 factor out a 2 in \[F(2*K, 2*L)\] \[(2) * (10 * 2^{1/3} * K + 2^{1/2} * L)\] given the exponents, that is obviously less than \[2*(10*2^{2/3}*K + 2^{1/2}*L)\]

LeoMessi
 2 years ago
Best ResponseYou've already chosen the best response.0yep, so decreasing returns to scale

LeoMessi
 2 years ago
Best ResponseYou've already chosen the best response.0right on, so it seems safe to assume that when the exponents are less than 1, decreasing returns are most likely, and when the exponents are 1 then constant, and if greater than 1 than increasing

LeoMessi
 2 years ago
Best ResponseYou've already chosen the best response.0cool stuff, thanks man

dpflan
 2 years ago
Best ResponseYou've already chosen the best response.3yeah, I think you get the idea, try out some more problems
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