Thread: Area and Perimeter

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.

I would begin by stating the area and perimeter functions. Can you state these?

area: 2x(8/1+x^2)

Perimeter: 4x+ 2(8/1+x^2)

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.

then what would you do after youve substituted it in?

Once we have the area and perimeter as a function of x alone, then equate the derivatives to zero to find the critical numbers. Use an appropriate test to ensure you have maximized the two functions.