A community for students.
Here's the question you clicked on:
 0 viewing

This Question is Closed

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

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

LeoMessi
 3 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
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.0yep, so decreasing returns to scale

LeoMessi
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.0cool stuff, thanks man

dpflan
 3 years ago
Best ResponseYou've already chosen the best response.3yeah, I think you get the idea, try out some more problems
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.