The Attempt at a Solution
First I decided to find the equivalent resistance.
Since r2 and r3 are parallel, this can be combined into a resistor of resistance r2r3/(r2 + r3). Since r2 is known, I went ahead and wrote that as 14r3/(14+r3).
Now, this resistor is in series with r1, so I can find the equivalent resistance to be
2 + [14r3/(14+r3)]
Now, I can call the current due to the battery
i = V/R = V/(2 + [14r3/(14+r3)])
Allowing me to use the dissipation formula for r3
P = r3[V/(2 + [14r3/(14+r3)])]^2
Now I figured it would be a matter of differentiating this function and setting it equal to zero, but the derivative is unreasonably complex.