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.

Printable View

- Dec 9th 2012, 11:27 PMcamjensonArea 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.

- Dec 9th 2012, 11:33 PMMarkFLRe: Area and Perimeter
I would begin by stating the area and perimeter functions. Can you state these?

- Dec 9th 2012, 11:36 PMcamjensonRe: Area and Perimeter
area: 2x(8/1+x^2)

Perimeter: 4x+ 2(8/1+x^2) - Dec 9th 2012, 11:43 PMMarkFLRe: Area and 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. - Dec 10th 2012, 12:08 AMcamjensonRe: Area and Perimeter
then what would you do after youve substituted it in?

- Dec 10th 2012, 12:15 AMMarkFLRe: Area and Perimeter
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.