For above circuit
$$
\begin{array}{l}
\frac{1}{R^{\prime}}=\frac{1}{R}+\frac{1}{X}=\frac{R+X}{R X} \\
R^{\prime}=\frac{R X}{R+X}
\end{array}
$$
The current in the circuit $i=\frac{E}{R+R^{\prime}}=\frac{E}{\left(R+\frac{R X}{R+X}\right)}$
$V_{k=}=\frac{E \cdot \frac{R X}{R+X}}{\left(R+\frac{R X}{R+X}\right)}=\frac{E X}{R+2 X}$
Power dissipated in the circuit
$\begin{aligned} P_{x} &=\frac{V_{B X}^{2}}{X}=\frac{E^{2} \cdot X^{2}}{X(R+2 X)^{2}} \\ &=\frac{E^{2} X}{(R+2 X)^{2}} \\ \frac{d P_{X}}{d X} &=E^{2} \frac{(R-2 X)}{(R+2 X)^{3}} \\ d P_{X} &=\frac{E^{2}(R-2 X)}{(R+2 X)^{3}} \cdot d X \end{aligned}$
$\left(d P_{x}\right)$ will be zero for all $(d X)$ if $X=\frac{R}{2} \quad\left[d P_{x}=\frac{E^{2}\left(R-2 \times \frac{R}{2}\right)}{\left(R+2 \times \frac{R}{2}\right)^{3}}\right]$
$$
=0