Homework question from economics

[beaf123]

< Mentor Note -- thread moved to HH from the technical math forums, so no HH Template is shown >

Hi, I have already delivered the homework so my reason for asking is because I want to/ need to understand it.

I am at question d. I have found the two differential equations.

They are:

c° = -1/0,5 * ( f'(k) - 0,04)C_t

and

k° = f(k)-0,06k_t -C_t

In equilibrium both c° and k° will be zero so both equations can be set equal to zero.

But how do you proceede from here, to find k* and c*? Can you set f'(k) = 0,25K^-0,75 + 2 since f(k) = k^0,25 and k(0) = 2?

If I have explained it too bad I can elaborate. I hope someone can help me:)

 

[Ray Vickson]

< Mentor Note -- thread moved to HH from the technical math forums, so no HH Template is shown >

Hi, I have already delivered the homework so my reqson for qsking is becquse I want to/ need to understand it.

View attachment 80248

I am at question d. I have found the two differential equations.

They are:

c° = -1/0,5 * ( f'(k) - 0,04)C_t

and

k° = f(k)-0,06k_t -C_t

In equilibrium both c° and k° will be zero so both equations can be set equal to zero.

But how do you proceede from here, to find k* and c*? Can you set f'(k) = 0,25K^-0,75 + 2 since f(k) = k^0,25 and k(0) = 2?

If I have explained it too bad I can elaborate. I hope someone can help me:)

The posted images are too small on my screen, and I cannot enlarge them. My eyes are not as good as they once were, so your images are unreadable to me, reading glasses and all. If you post larger images or---better still---type it all out (as per PF standards), then I will be willing to help.

 

[beaf123]

Thanks. I will post a lager picture tomorrow.

 

[beaf123]

I have answered until c),

question d) is:

d) Find the equilibrium (k* c*).

And its here I am struggeling. I have my two equations, and all the needed numbers, so all I want to do is solve the two equations for C and K.

 

[Ray Vickson]

< Mentor Note -- thread moved to HH from the technical math forums, so no HH Template is shown >

Hi, I have already delivered the homework so my reason for asking is because I want to/ need to understand it.

View attachment 80248

I am at question d. I have found the two differential equations.

They are:

c° = -1/0,5 * ( f'(k) - 0,04)C_t

and

k° = f(k)-0,06k_t -C_t

In equilibrium both c° and k° will be zero so both equations can be set equal to zero.

But how do you proceede from here, to find k* and c*? Can you set f'(k) = 0,25K^-0,75 + 2 since f(k) = k^0,25 and k(0) = 2?

If I have explained it too bad I can elaborate. I hope someone can help me:)

If $c^{o}$ means $dc/dt$, etc., and if your two differential equations are correct (I have not checked) then the answer to your questions is YES. You need $dc/dt = 0$, so if $c \neq 0$ you need $f'(k^*) = 0.04$. Thus, you can determine $k^*$. You also need $dk/dt = 0$, so $0 = f(k^*) - 0.06 k^* - c^*$, so you can determine $c^*$

 

[beaf123]

Thank you. I found a mistake in my differential equation. It should be:
c° = -1/0,5 * ( f'(k) - 0,10)Ct

So this gives:

f`(k) = 0,10
0,25k^-0,75=0,10
k^-0,75=0,4
k=0,4^(-1/0,75)
k*= 3,393
which again gives

c*= 3,393^0,25-0,06(3,393)
c*=1,153

Thanks again for the help. Would you get the same answer if you inserted c_t = f(k) - 0,06k_t into the first equation?

 

[Ray Vickson]

Thank you. I found a mistake in my differential equation. It should be:
c° = -1/0,5 * ( f'(k) - 0,10)Ct

So this gives:

f`(k) = 0,10
0,25k^-0,75=0,10
k^-0,75=0,4
k=0,4^(-1/0,75)
k*= 3,393
which again gives

c*= 3,393^0,25-0,06(3,393)
c*=1,153

Thanks again for the help. Would you get the same answer if you inserted c_t = f(k) - 0,06k_t into the first equation?

Try it for yourself!