What are the dimensions of such a rectangle with the greatest possible area?

Question

A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola y= 9-x^2. What are the dimensions of such a rectangle with the greatest possible area?

 

width= ?????

height=?????

Answer 1

9-x^2 = y

is a downward opening parabola with vertex =maximum point = (0,9)

maximize the area enclosed within the x-axis, y-axis and lines to to a 4th vertex (x,y) = (x,9-x^2) in quadrant I

Area = of the right half of the rectangle = xy = x(9-x^2) = 9x -x^3

A' = 9-3x^2 = 0

3x^2 =9

x^2 =9/3=3

x = sqr3

y = 9-3=6

dimensions of the full rectangle are 6 by 2sqr3

max area for an inscribed rectangle = 12sqr3