= Solution
At fixed <temperature> difference and constant internal resistance,
$$
P_L=I^2R_L=\frac{V_{\rm oc}^2R_L}{(R+R_L)^2}.
$$
Differentiation gives
$$
\frac{dP_L}{dR_L}=\frac{V_{\rm oc}^2(R-R_L)}{(R+R_L)^3}.
$$
It is positive below $R_L=R$ and negative above it, while <power> vanishes for an open or shorted load. Therefore the <maximum power transfer theorem> gives
$$
\boxed{R_L=R,\qquad P_{L,\max}=\frac{V_{\rm oc}^2}{4R}
=\frac{S_{\rm eff}^2\Delta T^2}{4R}.}
$$
At this point $I=V_{\rm oc}/(2R)$ and half the electromotive <force> drops across the load. Maximum electrical <power> is not generally maximum thermal-conversion efficiency. If changing current also changes the <temperature> difference because the thermal source has finite impedance, the fixed-$\Delta T$ optimization must be reconsidered.
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