Let z = x 2+2y, then y =-x 2/2+z/2. The image is a parabola with a downward opening (the intersection with the Y axis is (0, z/2), and this problem is converted into the maximum value of z when (0, z/2) is in the curve).

When 0

When b>2, the curve is an ellipse with the focus on the Y axis.

When b=2, the curve is a circle,

Drawing shows that z "2b, that is, the maximum value is 2b,

So choose (d).