The voltage is measured across the rod segment of resistance $\lambda h$. Therefore,
\[U_1 = \mathcal{E} \frac{\lambda h}{\lambda a + R_0}\]
The bridge circuit should be analyzed. Since the voltmeter internal resistance is high, the two branches can be treated indepedently. The potential difference accross the voltmeter is:
\[ U_2 = \mathcal{E} \left( \frac{h}{a} - \frac{R_1}{R_1 + R_2} \right) \]
A deviated value of $R_1/R_2$ shifts $V(U_2)$ but doesn't change the slope, so the difference between the maximum and minimum volumes remains $V_0$. From the graph,
$V_0=68~\mathrm{dm}^3+8~\mathrm{dm}^3=76~\mathrm{dm}^3$.
At the initial point, the actual volume of fuel is $V=0$ but we observe that
\[ -\frac{2}{19} = \frac{U_{2,V=0}}{\mathcal{E}} + \frac{1}{2}, \quad \Rightarrow \quad \frac{U_{2,V=0}}{\mathcal{E}} = -\frac{23}{38}.\]The correct value of $R_1/R_2$ satisfies the condition:
\[ \frac{23}{38} = \frac{R_1/R_2}{R_1/R_2 + 1} \quad \Rightarrow \quad R_2/R_1 +1 = \frac{38}{23}, \quad \Rightarrow R_2/R_1 = \frac{15}{23}\]
The friction power $P = \beta v^2$ must equal the mechanical power $\eta P_0$ of the engine. Also, $v = \omega_W r = k \omega r$.
\[\alpha P_0 \frac{\omega}{\omega_0} \left(1 - \frac{\omega}{\omega_0} \right) = \beta k^2 \omega^2 r^2\]After canceling $\omega$, we obtain a linear equation with the solution
\[\omega = \omega_0 \frac{1}{1+\frac{\beta k^2 r^2 \omega_0^2}{\alpha P_0}}\]
Instead of maximizing of $v$ we can minimize $\omega_0 r/v$:
\[\frac{\omega_0 r}{v} = \frac{1}{k} + k \frac{\beta r^2 \omega_0^2}{\alpha P_0}\]The derivative with respect to $k$:
\[ 0 = -\frac{1}{k_\text{max}^2} + \frac{\beta r^2 \omega_0^2}{\alpha P_0}\]