Let's prove that if you want to maximise \(ab\) with \(a + b\) equal to a constant value \(k\), then you want \(a = b = \frac{k}{2}\).

Take \(s = k/2\). If \(a = s+h\) then \(b = s-h\), from which we get that \(ab = (s+h)(s-h) = s^2 - h^2\). Because we know \(h^2 \geq 0\), \(ab\) is maximised when \(h = 0\), that is \(a = b = s = k/2\).

Twitter proof:

— Mathspp (@mathsppblog) November 5, 2020

Take s = k/2. If a = s+h then b = s-h, from which we get that ab = (s+h)(s-h) = s^2 - h^2. Because we know h^2 >= 0, ab is maximised when h = 0, that is a = b = s = k/2.https://t.co/9ypegSMF2F