## Algebra and Trigonometry 10th Edition

$x=\frac{S}{2}$ $y=\frac{S}{2}$ $P=\frac{S^2}{4}$
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. $y=S-\frac{S}{2}=\frac{S}{2}$