It would help you if you made a sketch.
If so, then it should be obvious that the x-intercepts are ±3
Let the point of contact in quadrant I be (x,y)
then the area of the rectangle is:
Area = 2xy
= 2x(9 - x^2) = 18x - 2x^3
d(Area)/dx = 18 - 6x^2 = 0 for a max/min of Area
6x^2 = 18
x^2 = 3
x = ± √3
Finish it up from here, it's a cakewalk.
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 =
1 answer