The area of the rectangle with corners (-x,0), (x,0) (-x,y) (x,y) is
a = 2x * (11-x^2)
a = 22x - 2x^3
da/dx = 22 - 6x^2
da/dx = 0 when x = √(11/3)
So, the rectangle is 2√(11/3) x 22/3 with area 44/3 * √(11/3) = 28.08
A rectangle is inscribed with its base on the x -axis and its upper corners on the parabola
y= 11-x^2.
What are the dimensions of such a rectangle with the greatest possible area?
1 answer
1 answer