If the base is 2x, then the height is 11-x^2, right? So
A = 2x(11-x^2)
dA/dx = 22-6x^2
so the max area is where x = √(11/3) and y = 22/3
That area is 44/3 √(11/3)
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