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        Wanting the Thevenin Eq. of the circuit external of R3 (= 2kΩ), the R3 is removed from the circuit. VTh becomes the open circuit voltage at the terminals, which is 75V. 
        For RTh, we can treat current sources as open circuits, voltage sources as short circuits, and add Rs in parallel and series. Adding R1 and R2 in parallel gives RTh = 5.33kΩ.

        Alternatively, we could use Norton's theory to treat the open circuit as a short circuit and find the current through it. Using Kirchoff''s current law, I3= 11.81 + 2.25 ~ 14mA. Using IN(I3) = VTh/RTh, RTh = 5.33Ω.

              VTH and RTH is found the same way as part 1. The voltage across the open circuit approximates to 61.2V, and adding R1 and R2 in parallel gives the RTh = 42.4Ω.
              If the R of the load is to dissipate as much power as possible, it must be equivalent to RTh; therefore, RL = 42.4. Using PL = VTh/4RTh, PL = 22.1W.