For y = 2, . Note that is always negative.(B) If S represents the square of the distance from (3, 0) to a point (x, y)
on the curve, then S = (3 − x)^2 + y^2 = (3 − x)^2 + (x^2 − 1). Setting
yields the minimum distance at .(D)(D) See the figure. Since the area, A, of the ring equals ,and .It can be verified that x = 2 produces the maximum area.