# Thread: Basic but frustrating maximisation problem

1. ## Basic but frustrating maximisation problem

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.

2. ## Re: Basic but frustrating maximisation problem

Originally Posted by frustrated

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

3. ## 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.

4. ## 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 !

5. ## Re: Basic but frustrating maximisation problem

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

6. ## Re: Basic but frustrating maximisation problem

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

7. ## 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 ?

8. ## Re: Basic but frustrating maximisation problem

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

9. ## 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.