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.