Update
The following seems to give a good approximation for the last problem but I don't know why:
X= 1-1/[(a-b)(a-1)+2] as a-b -> 0 for small values of a-b the solution is almost perfect
We need to find the value of X that maximizes U,
U= [(w*X)^(1-a)-1]/(1-a)*[(s*(1-X))^(1-b)-1]/(1-b)
where 0<X<1 w>1 s>1
For special cases of a=1, we have as the first term Ln(wX); for b=1 we have Ln(s(1-X)
We are allowed to assume w and s are large such that both terms are positive
Taking the first differential using the product rule I get:
dU/dX= 0= w(w*X)^-a * [(s*(1-X))^(1-b)-1]/(1-b) + s(s*(1-X))^-b *[(w*X)^(1-a)-1]/(1-a)
Which doesn't really seem to get me very far!
Any idea of alternative approaches or next steps ?
We can get a neater solution under special conditions w=S -> infinity; a>1; b>1, which are acceptable approximations
0 = [X^-a / (b-1)] + [(1-X)^-b / (a-1)]
X^-a / (1-b) = -(1-X)^-b / (1-a)
X^-a * (a-1) = (1-X)^-b *(b-1)
But even here I am still stuck on how to proceed.
Update
The following seems to give a good approximation for the last problem but I don't know why:
X= 1-1/[(a-b)(a-1)+2] as a-b -> 0 for small values of a-b the solution is almost perfect