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

maximising ab with a+b=k is achieved with a=b=k/2

Twitter proof

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 \(a = b = s = k/2\).

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