The maximum area (in sq. units) of a rectangle having its base on the x-axis and its other two vertices on the parabola, y = 12 -- x2 such that the rectangle lies inside the parabola, is :
Solutions · 1
Official solution
f (a) = 2a(12 - a)2
f '(a) = 2(12 - 3a2)
Maximum at a = 2
maximum area = f(2) = 32