Thread: 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.

Originally Posted by frustrated

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 rather than , and put , also assume . Then your problem is to find the that maximises:



so:



so for an extremum you need to solve , which if is equivalent to:



Note: The maximising in your notation cannot depend on 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 which gives this.


CB

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.

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 !

Originally Posted by frustrated

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

Originally Posted by frustrated

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

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 ?

Originally Posted by frustrated

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

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.