F(k,l)=K(squared) + L(squared)

SOMEONE PLEASE HELP---

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- Sep 17th 2007, 07:47 PMnartyhomogenous
F(k,l)=K(squared) + L(squared)

SOMEONE PLEASE HELP--- - Sep 17th 2007, 08:10 PMnartyintroduce 'z'
to the equation and solve---

the one i could do was---

F(k,l)='square root'of kl

F(zK,zL)=Z'square root of K and L' - Sep 17th 2007, 08:57 PMnartyHheeellpppp..
anyone!!!

- Sep 17th 2007, 09:07 PMAfterShock
Uh, I'm not sure what exactly you're wanting...

By your insertion of the "Z"s, this is simply showing that this is a homogeneous function of degree 2:

$\displaystyle f(\lambda k, \lambda l) = (\lambda k)^2 + (\lambda l)^2 = \lambda^2 (k^2 + l^2) = \lambda^2\cdot f(k,l)$ - Sep 17th 2007, 09:16 PMnarty
THANK YOU GREATLY------AND what would my other problem be---

F(K,L)= 'square root' of K + 'square root' of L - Sep 17th 2007, 09:21 PMnartythe first problem i did was this----
F(K,L) = 'square root' of KL

= 'square root' of zK times zL

= z times the square root of KL

= zF(K,L), a constant return to scale - Sep 17th 2007, 09:27 PMAfterShock
In general, $\displaystyle f(\lambda t, \lambda y) = \lambda^\alpha f(t,y)$ for some real number $\displaystyle \alpha \implies$ f is a homogeneous function of degree $\displaystyle \alpha \in \mathbb{R}$.

$\displaystyle f(\lambda k, \lambda l) = (\lambda k)^\frac{1}{2} + (\lambda l)^\frac{1}{2} = \lambda^\frac{1}{2}(k^\frac{1}{2} + l^\frac{1}{2}) = \lambda^\frac{1}{2}\cdot f(k,l)$. - Sep 17th 2007, 09:30 PMnartyso,
AND what would my other problem be---

F(K,L)= 'square root' of K + 'square root' of L

------with introducing 'z' to solve?

- Sep 17th 2007, 09:37 PMAfterShock
- Sep 17th 2007, 09:42 PMnartyim having problems doing that..
----ALSO is this algebra of calculus-------I dont even know that----AND thanks GREATLY for all your help--cause i need to know these by 8am today-----__How would you do it--cause mine doesnt look right__ - Sep 17th 2007, 09:50 PMAfterShock
I've shown you the general method to determine whether a function is homogeneous, and if it is of what degree. It's hard trying to figure out if this is truly what you're wanting, as you never state the question.

I'd like to emphasize that you will never understand this material if you rely on someone else to do*all*the work for you. - Sep 17th 2007, 09:54 PMnarty
he just said to introduce 'z' (which is greater than one) and solve(by introducing 'z')------yeah i know i have to learnthis stuff but now its like chinese----i got someone helping me next week with a refresher course--but now i have to do these problems somehow and dint know where to begin, so ANY help is greatly appreciated--THANKX---:)