RE: Maximum of a rational function of many variables in the unit cube

[Jacques Gélinas <jacquesg00@hotmail.com>]

Thanks for the suggestions, including forvec() and Lagrange multipliers.

It turns out that I can find the maximum in the unit hypercube for 

these special expressions by eliminating one variable at a time.

If 0<a<1, 0<b<1, and 0<c<1, then

a/3 = 1/3 - (1-a)/3 < 1/3   

1/2 * (1-a*b/5) / (1-a/3) = -1/2*[a*b/(1-a/3) + 3/2(1-2/(3-a)] + 3/4 < 3/4

f2(a,b,c) = 4/5 - [  ..... ] < 4/5

GP is essential here to check the algebra.

Jacques Gélinas