Hallo cb,
G(q) = E(q) - K(q)
= 722*q - ( 0.065 q3 - 3.9514 q2 + 505 q + 7700)
≈ - 0.065·q3 + 3.9514·q2 + 217·q - 7700
G '(q) = - 0.195·q2 + 7.9028·q + 217 = 0
→ q = 59.2948 [ ∨ q = -18.76759755 ] gewinnmaximale Menge in Mbbl
K(59.2948) ≈ 37302 GE
37302 GE / 21 = 1776.286 GE pro Plattform
Gruß Wolfgang