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Homework Help: Homework question from economics

  1. Mar 12, 2015 #1
    < 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)Ct


    k° = f(k)-0,06kt -Ct

    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:)
    Last edited: Mar 12, 2015
  2. jcsd
  3. Mar 12, 2015 #2

    Ray Vickson

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    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.
  4. Mar 12, 2015 #3
    Thanks. I will post a lager picture tomorrow.
  5. Mar 13, 2015 #4

    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.
    Last edited: Mar 13, 2015
  6. Mar 13, 2015 #5

    Ray Vickson

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    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^*##
  7. Mar 14, 2015 #6
    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
    k*= 3,393
    which again gives

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

    Thanks again for the help. Would you get the same answer if you inserted ct = f(k) - 0,06kt into the first equation?
  8. Mar 14, 2015 #7

    Ray Vickson

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    Try it for yourself!
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