# Cobb-Douglas Function - Help obtaining k*

• Oct 8th 2012, 12:16 PM
entrepreneurforum.co.uk
Cobb-Douglas Function - Help obtaining k*
Can somebody walk me through why this holds from the cobb douglas function AK(t)^α - L(t)^1-α

I don't know how to obtain the steady state s(Ak(t)^α) = δk(t) from the function above
And I want to know how to get k at equilibrium k* => k* =
(​δ/sA)^(1/1+&)

Thanks

Edit: I cannot see to fix the TEX error, sorry
• Oct 8th 2012, 07:43 PM
chiro
Re: Cobb-Douglas Function - Help obtaining k*
Hey entrepreneurforum.co.uk.

Steady states are typically obtained by solving for the derivative to be equal to zero. It looks like you only have one parameter t, so differentiate the function, set it zero and solve for the particular value.

Do you have a differential equation form for the function? Also can you point out to us what the δk(t) refers to?

Typically a steady-state solution deals with a situation where you have an implicit differential equation where the changes depend only on the values that just preceded them (but not always).
• Oct 9th 2012, 12:09 AM
entrepreneurforum.co.uk
Re: Cobb-Douglas Function - Help obtaining k*
Hello,

δk(t) = the depreciation rate of (k) capital
so far I've do done the following;

Y(t) = AK(t)^α - L(t) ^(1-α)

Divided through by L to obtain (little) y (as output per unit of labour)

y(t) = AK(t) ^α = f(k)

and then finally get the steady state which would be;

k(t+1) = s(Ak(t)^α) = δk(t)

But I don't know the last steps..

Also regarding k* we use the steady state function to obtain.

ALSO: t is not a parameter its just a reference to (time at that point)
• Oct 9th 2012, 12:20 AM
chiro
Re: Cobb-Douglas Function - Help obtaining k*
Try isolating the k* (you have written k(t)) term and get it in terms of the other variables.