I've just done a bit more thinking. If I've got my equation right, and calling the extra distance the ship sails on "x" (nm), then there IS indeed a benefit, but it is vanishingly small.
Time taken for ship to return to point of release is (x/30)+((300-x)/150). Boiling that down a bit, I get (2x+150)/75. Differentiating that gets me to 2/75ths of a nautical mile, or probably less than the turning circle of the ship!
Or have I stuffed up my algebra/calculus? It's been a while...