Results 1 to 6 of 6

Math Help - Utility functions

  1. #1
    Newbie
    Joined
    Feb 2011
    Posts
    15

    Utility functions

    Hi!

    I've got the following utility function:

    U = (y_1 - 2) (y_2 - 4) = y_1y_2 - 4y_1-2y_2-8

    prices are

    p_1 = 2 , p_2=4

    p_1 is the price of good 1
    p_2 is the price of good 2

    y_1 is the quantity of good 1
    y_2 is the quantity of good 2

    The budget costraint is:

    100 = 2y_1 + 4y_2

    Now, we know that we have equilibrium when we have

    \frac { \delta U / \delta y_1} {p_1} = \frac { \delta U / \delta y_2} {p_2}

    because weighted marginal utilities are equal.

    Ok, now this is where i get stuck! The book "says" that

    with our data, we have:

    \frac { \delta U / \delta y_1} {p_1} = \frac {y_2 - 4} {2} and \frac { \delta U / \delta y_2} {p_2} = \frac {y_1 - 2} {4}
    My question is: why are \delta U / \delta y_1 equal to y_2 - 4 and \delta U / \delta y_2 equal to equal to y_1 - 2 ?

    Thank you in advance!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    May 2010
    Posts
    1,030
    Thanks
    28
     \displaystyle \frac{\delta U}{\delta y_1} is the partial differential of U with respect to y1. This means you differentiate the function with respect to y1, treating all other variables as constants

     \displaystyle \frac{\delta U}{\delta y_1}=\frac{\delta (y_1 y_2)}{\delta y_1} + \frac{\delta (-4 y_1) }{\delta y_1} + \frac{\delta ( -2 y_2) }{\delta y_1} + \frac{\delta (-8)}{\delta y_1}

    Taking the first term, remembering to treat y2 as a constant when differentiating, this is
     \displaystyle \frac{\delta (y_1 y_2)}{\delta y_1} = y_2



    Now all the other terms
     \displaystyle \frac{\delta U}{\delta y_1}=\frac{\delta (y_1 y_2)}{\delta y_1} + \frac{\delta (-4 y_1) }{\delta y_1} + \frac{\delta ( -2 y_2) }{\delta y_1} + \frac{\delta (-8)}{\delta y_1}

     \displaystyle \frac{\delta U}{\delta y_1}= y_2 - 4 + 0 + 0

     \displaystyle \frac{\delta U}{\delta y_1}= y_2 - 4

    You try the other one.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Feb 2011
    Posts
    15
     \displaystyle \frac{\delta U}{\delta y_2}=\frac{\delta (y_1 y_2)}{\delta y_2} + \frac{\delta (-4 y_1) }{\delta y_2} + \frac{\delta ( -2 y_2) }{\delta y_2} + \frac{\delta (-8)}{\delta y_2}

     \displaystyle \frac{\delta (y_1 y_2)}{\delta y_2} = y_1

     \displaystyle \frac{\delta (-4 y_1) }{\delta y_2} = 0 (because y_1 is a constant and its derivative is 0, multiplied by -4 gives 0, right?)

     \displaystyle \frac{\delta ( -2 y_2) }{\delta y_2} = -2 (because y_2's derivative is 1, multiplied by -2 gives -2, right?)

     \displaystyle \frac{\delta (-8)}{\delta y_2} = 0 (because 8 is a constant and its derivative is 0?)

     \displaystyle \frac{\delta U}{\delta y_2} = y_1 - 2

    Ok, thank you. Please, tell me if I did something wrong!

    Thank you again!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    May 2010
    Posts
    1,030
    Thanks
    28
    perfect!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Feb 2011
    Posts
    15
    Quote Originally Posted by SpringFan25 View Post
    perfect!
    Ok, thanks again
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Feb 2011
    Posts
    15
    Hi, again. I've got another problem here. I have to maximize the following Utility function: max U = U(y_1, y_2)

    subject to the following budget costraint:  R = y_1 p_1 + y_2 p_2

    Of course \displaystyle y_1=\frac {R - y_2 p_2} {p_1}

    Then max U = U(\frac {R - y_2 p_2} {p_1}, y_2)

    Clear till here... now he says:

    By differentiating this expression and making it equal to zero, we get

    \displaystyle \frac {\delta U}{\delta y_1} \frac {\delta y_1}{\delta y_2} +\frac {\delta U}{\delta y_2} = 0
    Of course I don't want to take advantage of your helpfulness again, but could you please just tell me if I have to do the same things we've seen in my first post? Thanks again!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Utility question
    Posted in the Business Math Forum
    Replies: 1
    Last Post: August 5th 2010, 10:38 AM
  2. Expected utility.
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: May 1st 2010, 08:42 PM
  3. Utility functions and insurance
    Posted in the Business Math Forum
    Replies: 1
    Last Post: August 23rd 2009, 07:14 AM
  4. More Utility Functions
    Posted in the Business Math Forum
    Replies: 0
    Last Post: August 18th 2009, 07:37 AM
  5. Graphing utility
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: January 22nd 2008, 01:14 AM

Search Tags


/mathhelpforum @mathhelpforum