Answer
$x=12$
$y=6$
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 $y$
$x+2y=24$
$x=24-2y$
Product:
$P=xy$
$P=(24-2y)y=24y-2y^2$
$P=-2(y^2-12y)$
$P=-2[(y^2-2(6)y+6^2)-6^2]$
$P=-2(y-6)^2+72~~$ (Notice that: $a=-2$. Parabola opens downward)
So the vertex $(6,72)$ is the maximum.
$x=24-2(6)=12$
