homogenous

**narty** · September 17th 2007, 07:47 PM

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

SOMEONE PLEASE HELP---

**narty** · September 17th 2007, 08:10 PM

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'

**narty** · September 17th 2007, 08:57 PM

anyone!!!

**AfterShock** · September 17th 2007, 09:07 PM

Originally Posted by narty

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

SOMEONE PLEASE HELP---

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:

$f(\lambda k, \lambda l) = (\lambda k)^2 + (\lambda l)^2 = \lambda^2 (k^2 + l^2) = \lambda^2\cdot f(k,l)$

**narty** · September 17th 2007, 09:16 PM

THANK YOU GREATLY------AND what would my other problem be---

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

**narty** · September 17th 2007, 09:21 PM

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

**AfterShock** · September 17th 2007, 09:27 PM

Originally Posted by narty

THANK YOU GREATLY------AND what would my other problem be---

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

In general, $f(\lambda t, \lambda y) = \lambda^\alpha f(t,y)$ for some real number $\alpha \implies$ f is a homogeneous function of degree $\alpha \in \mathbb{R}$ .

$f(\lambda k, \lambda l) = (\lambda k)^\frac{1}{2} + (\lambda l)^\frac{1}{2} = \lambda^\frac{1}{2}(k^\frac{1}{2} + l^\frac{1}{2}) = \lambda^\frac{1}{2}\cdot f(k,l)$ .

**narty** · September 17th 2007, 09:30 PM

AND what would my other problem be---

F(K,L)= 'square root' of K + 'square root' of L
------with introducing 'z' to solve?

**AfterShock** · September 17th 2007, 09:37 PM

Originally Posted by narty

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

$f(k,l) = (kl)^\frac{1}{2}$

Try plugging in $\lambda$ and you'll notice $f(\lambda k, \lambda l) = \ldots$

It's trivial. Work it through just like the others.

**narty** · September 17th 2007, 09:42 PM

How would you do it--cause mine doesnt look right----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-----

**AfterShock** · September 17th 2007, 09:50 PM

Originally Posted by narty

How would you do it--cause mine doesnt look right----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-----

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.

**narty** · September 17th 2007, 09:54 PM

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---

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homogenous

introduce 'z'

Hheeellpppp..

the first problem i did was this----

so,

im having problems doing that..

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