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If the objective is to maximize the power dissipated on the load \$R\$. Use,

\$P_R = \frac{U^2 \cdot R}{(R + 3)^2 + 4^2} = \frac{U^2 \cdot R}{ R^2+6R+9+16} = \frac{U^2}{ R+6+25/R}\$ Then, derive the expression to find its maximum, that is, when \$R+6+25/R\$ reaches its minimum.

If you are actually looking for maximizing the power dissipated on both resistors, \$R\$ and \$ 3 \Omega \$, use,

\$P_R = \frac{U^2 \cdot (R+3)}{(R + 3)^2 + 4^2}\$

jDAQ
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