A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola y= 6-x^2. What are the dimensions of such a rectangle with the greatest possible area?

Answer 1

x^2 - 6 = - y
vertex on y axis,
opens down (sheds water)
so just do the right half
half A = x y
hA = -x (x^2-6) = -x^3 + 6 x
dhA/dx = 0 at max = -3 x^2 + 6
x^2 = 2
x = sqrt 2
include left half
length along x axis = 2 sqrt 2
if x = sqrt 2
y = 6 - x^2 = 6-2 = 4

Answer 2

thank you, somebody wad telling me you had to find the derivative first but i kept getting wrong answers from it