I have an optimization problem in the form max \((a\bar{a})(b\bar{b})\) subject to \(a+b=1\) Here \(\bar{a}\) and \(\bar{b}\) are known values and both of them are >0. Let \(a_{opt}\) and \(b_{opt}\) are the solutions or optimal values then \(a_{opt}\bar{a}>0\) and \(b_{opt}\bar{b}>0\) How to find this solution for this optimization problem? asked 09 Feb '16, 04:13 dimitrios 
As Rob suggests, substitute to get a function of one variable (a). Then first derivative to find turning point gives optimal a (second derivative will show it's a maximum). Substitute back to get b_opt. You should find a = (1 + a¯  b¯)/2, Also, your last set of constraints on the optimal values constrains the sum of a¯ + b¯ < 1 answered 09 Feb '16, 11:55 Grallator @Grallator, thanks for the answer!
(09 Feb '16, 20:35)
dimitrios

How big could be those \(a\) and \(b\)?
Substitute \(b=1a\) into your objective function, yielding a quadratic polynomial in \(a\). To find \(a_\text{opt}\) and \(b_\text{opt}\), use either singlevariable calculus or a wellknown fact about the location of a parabola's vertex with respect to its intercepts.
@Rob Pratt, Thank you very much for the hints, i.e., the solutions!