if the rectangle extends from -x to x,
a = 2xy = 2x(8-x^2)
da/dx = 16 - 6x^2
a has a max at x = 4/√6
width = 8/√3
height = 16/3
A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola y = 8−x2. What are the dimensions of such a rectangle with the greatest possible area?
width=
length=
1 answer