I think I understand both examples. I don't think either refutes my
assertion that the weight ends up at the same place either way so work
delivered is the same. The "area under the path" argument works out
the same in both cases. Delivered work is the total available energy
(area of the whole triangle ( 1/2 * 30 * 2 = 30 in-lb ) minus the
remaining energy which is the area of the rectangle (15 * 1 = 15
in-lb) or 15 in-lb delivered.