Minimization sketch: keep turnover constant (Q×hours = const). With Q ∝ RPM and P ∝ RPM³, energy/day E = P×hours ∝ RPM³ × (const/RPM) = const × RPM². So the cheapest schedule is the lowest feasible RPM that still achieves the same turnover in ≤24 hours and within the pump’s allowed RPM range.