Let f(p) be profitability as a function of the parameter setting p. Suppose that a parameter of 0 leads to 0 profitability, and that f is always positive. Suppose also that the best value is as far as possible from 120. Near a maximum, profitability is well-approximated by
f(120) ~= f(115) + f'(115) * 5 + f''(115) * 5^2 / 2.
~= f(115) + f''(115) * 5^2 / 2.
Note now that f''(115) is probably no more than 500, assuming no great concavity changes, since
0 < f(0) ~= f(120) + f''(120) * 120^2 / 2
so f(120) ~= $5M,
f''(120) ~<= $10M / 120^2 = 1000.
This means that
f(120) - f(115) <= 1000 * 12.5.
So the difference in profitability is around $12500. Therefore, this is probably not worth working on.