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Math Help - Basic but frustrating maximisation problem

  1. #1
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    Basic but frustrating maximisation problem

    Please help me with this basic problem that I cannot solve;

    Consider the following equation

    U=(H-h) * (I^(1-e))/1-e

    and the following identity

    I=h*W

    and the following condition

    H>h

    I want to solve for e when U is at its maximum. By guess is to transform to

    U=(H-h) * ((hw)^(1-e))/1-e

    and differentiate wrt to h , then set dU/dh to zero and solve. This method should work as it is an inverted U shaped function.

    or alternatively express as I

    U=(H-I/w) * (I^(1-e))/1-e

    and differentiate wrt to I.

    The expected result is e=H/(H-h), but I cannot do the proof !

    Thanks in advance, my maths is very rusty, and i cannot get any further, unfortunately.
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  2. #2
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    Re: Basic but frustrating maximisation problem

    Quote Originally Posted by frustrated View Post
    Please help me with this basic problem that I cannot solve;

    Consider the following equation

    U=(H-h) * (I^(1-e))/1-e

    and the following identity

    I=h*W

    and the following condition

    H>h

    I want to solve for e when U is at its maximum. By guess is to transform to

    U=(H-h) * ((hw)^(1-e))/1-e

    and differentiate wrt to h , then set dU/dh to zero and solve. This method should work as it is an inverted U shaped function.

    or alternatively express as I

    U=(H-I/w) * (I^(1-e))/1-e

    and differentiate wrt to I.

    The expected result is e=H/(H-h), but I cannot do the proof !

    Thanks in advance, my maths is very rusty, and i cannot get any further, unfortunately.
    First change your notation, what you have makes it difficult to read and will result in some confusing notation.

    Let the free variable be x=1-e rather than e, and put k=H-h>0, also assume I>0. Then your problem is to find the x that maximises:

    U(x)= k \frac{e^{\ln(I)(x)}}{x}

    so:

    U'(x)=k\left[ \frac{\ln(I) e^{\ln(I)x}}{x}-\frac{e^{\ln(I)x}}{x^2}\right]

    so for an extremum you need to solve U'(x)=0, which if x\ne 0 is equivalent to:

    x=\frac{1}{\ln(I)}

    Note: The maximising e in your notation cannot depend on H as this only appears in the multipler which is by definition a constant greater than zero and so effects the maximum but not the value of e which gives this.


    CB
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  3. #3
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    Re: Basic but frustrating maximisation problem

    Thanks, Cptn Black, but i think it is wrong! You cannot treat K as a constant here becuase h=I/W, hence your K=H-I/W. The K contains the I term, so it is not a constant.
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  4. #4
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    Re: Basic but frustrating maximisation problem

    I think i have it solved;

    firtsly expand the first section to

    U= H* (I^1-e)/(1-e) - I/W * (I^1-e)/(1-e)

    bring the I in the second term up;

    U= H *1^1-e/(1-e) - I^2-e/(w*(1-e)

    differentiate wrt to I

    U' = H I ^ e - (2-e)/(w*(1-e))*I^1-e

    set U'to zero;

    H I ^ e = (2-e)/(W*(1-e)) * I^1-e

    divide both terms by I^e

    H=(2-e)(W*(1-e)) *I

    substitute h/W for I

    H= (2-e)(1-e)*h

    (2-e)(1-e) = H/h

    I think that is right !
    Last edited by frustrated; August 8th 2011 at 06:40 AM. Reason: minor error
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  5. #5
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    Re: Basic but frustrating maximisation problem

    Quote Originally Posted by frustrated View Post
    Thanks, Cptn Black, but i think it is wrong! You cannot treat K as a constant here becuase h=I/W, hence your K=H-I/W. The K contains the I term, so it is not a constant.
    Unless you are saying that W is an unspecified function of x then that is irrelevant (and if you are then we must assume that h and H are also potentially functions of x).

    CB
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  6. #6
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    Re: Basic but frustrating maximisation problem

    Quote Originally Posted by frustrated View Post
    I think i have it solved;

    firtsly expand the first section to

    U= H* (I^1-e)/(1-e) - I/W * (I^1-e)/(1-e)

    bring the I in the second term up;

    U= H *1^1-e/(1-e) - I^2-e/(w*(1-e)

    differentiate wrt to I

    U' = H I ^ e - (2-e)/(w*(1-e))*I^1-e

    set U'to zero;

    H I ^ e = (2-e)/(W*(1-e)) * I^1-e

    divide both terms by I^e

    H=(2-e)(W*(1-e)) *I

    substitute h/W for I

    H= (2-e)(1-e)*h

    (2-e)(1-e) = H/h

    I think that is right !
    Please in future identify which symbols represent the variables and which you think are constants or functions. Otherwise we are wasting our time looking at your questions.

    CB
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  7. #7
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    Re: Basic but frustrating maximisation problem

    Sorry, my bad. I should have made it clear that W and H are constants, hence, to specify I is to specify h. In this equation h is hours worked and w is wages per hour, and I is income, hence I=hw. I will try to be clearer in the future.

    The problem is that I have not done any proper maths for ten years, so I am really out of my depth.

    On a differnt note, is it possible to to further simplify the answer given above, assuming it is correct ?
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  8. #8
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    Re: Basic but frustrating maximisation problem

    Quote Originally Posted by frustrated View Post
    Sorry, my bad. I should have made it clear that W and H are constants, hence, to specify I is to specify h. In this equation h is hours worked and w is wages per hour, and I is income, hence I=hw. I will try to be clearer in the future.

    The problem is that I have not done any proper maths for ten years, so I am really out of my depth.

    On a differnt note, is it possible to to further simplify the answer given above, assuming it is correct ?
    And what is e? The base of natural logarithms?

    CB
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  9. #9
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    Re: Basic but frustrating maximisation problem

    e in this equation is an inequality/risk aversion paramater for the isoelastic utility from income function U=I^1-e/1-e. H is hours available to work or engage in leisure. The H-h term is thus lesure time. There is a straight multiplication of power-discounted income and leisure to give utility ,U

    It thus a special form of a Cobb-Douglas function

    U = L^a * I^b

    where in this instance a=1 and b=1-e so there is 'economies of scale' as a+b>1 when b is positive

    I just realised an easier way to perform this type of operation is to take the natural log of the equation, and maximise that instead.

    According to a reference book this leads to the generalised result

    h/H =b/(b+a)

    which from my specific equation form should give

    h/H = (1-e)/(2-e)

    I have checked the working, and actual numerical results calculated, and
    h/H =(1-e)/(2-e) is correct.

    I should have used another letter for epsilon !

    Working is as above, except that I made a minor typing error leaving out a divisor, proper result is below;

    U=(H-I/w) * (I^1-e)/1-e

    expand to

    U= H(I^1-e)/1-e - I/W(I^1-e)/1-e

    Bring the I on the right up and w down

    U= H(I^1-e)/1-e - (I^2-e)/w(1-e)

    differentiate wrt I gives

    Du/DI = HI^-e - (2-e)/w(1-e) * I^(1-e)

    Du/Di = 0

    HI^-e = (2-e)/w(1-e) * I^(1-e)

    divide by I^-e

    H= (2-e)/w(1-e) * I

    substitute for I =wh

    H=(2-e)/(1-e) *h

    H/h =(2-e)/(1-e)

    b= 1-e

    h/H = b/(1+b)

    If someone can explaint the log method under the budget constraint, that would be cool.

    apparetly we should have

    Log U(I,L) =ylnI +(1-y) ln L

    where y=b/(b+a)

    which under the budget constraints

    I = hw
    L=H-h
    L=H-(I/w)

    and the maximum for Log U(I,L) gives;

    h/H =y

    Cheers.
    Last edited by frustrated; August 8th 2011 at 05:49 PM. Reason: working added
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