Part a)

Determine R_o so that the maximum possible power is delivered to the 6 W resistor.

First, the power absorbed by the 6 W resistor is

P = V₆² / 6

By voltage division,

V₆ = 30 (6) / (6 + R_o) = 180 / (6 + R_o)

So

In order to maximize the power, the denominator should be minimized (since the numerator is constant).

Hopefully it is evident that setting R_o = 0 W will minimize the denominator, thus maximizing the power.

Note that this result is actually rather obvious, since any resistance inserted between the source and the 6 W resistor will reduce the current through the 6 W resistor, thus reducing the power.

Part b)

The maximum power that can be delivered to the 6 W resistor is

P = 30² / 6

P = 150 W