Answer
$x=\frac{S}{2}$
$y=\frac{S}{2}$
$P=\frac{S^2}{4}$
Work Step by Step
Quadric function in standard form:
$y=a(x-h)^2+k$, in which $(h,k)$ is the vertex. And, the maximum (or the minimum) occurs at the vertex.
Two positive real numbers: $x$ and $y$
$x+y=S$
$y=S-x$
Product:
$P=xy$
$P=x(S-x)=Sx-x^2=-x^2+Sx$
$P=-(x^2-Sx)$
$P=-[(x^2+2(\frac{S}{2})x+(\frac{S}{2})^2)-(\frac{S}{2})^2]$
$P=-[x-(\frac{S}{2})^2]+\frac{S^2}{4}~~$ (Notice that: $a=-1$, so the parabola opens downward)
So the vertex $(\frac{S}{2},\frac{S^2}{4})$ is the maximum.
$y=S-\frac{S}{2}=\frac{S}{2}$
