Rectangle in first quadrant has one vertex at origin and opposite vertex on the curve y=8/(1+x^2). Show the rectangle with max area isn't the same as the rectangle with max perimeter.
One vertex is at (0,0) and the other is at (x,y) (on the given curve) where x and y are positive. So we have:
Area: xy
Perimeter: 2(x+y)
Do you see that you have doubled x? Now substitute for y using the definition of the given curve.