= Solution
In the $(p,V)$ plane, the <Otto cycle> consists of:
* $A\to B$: an adiabatic compression along $pV^\gamma=\text{constant}$ from $V_1$ to $V_2$;
* $B\to C$: a vertical constant-volume rise at $V_2$;
* $C\to D$: an adiabatic expansion back to $V_1$;
* $D\to A$: a vertical constant-volume fall at $V_1$.
It is a clockwise loop, so its enclosed area is the positive net work output. In the $(T,S)$ plane, the two reversible adiabats $A\to B$ and $C\to D$ are vertical constant-entropy segments. Constant-volume heating $B\to C$ moves upward and to the right, while constant-volume cooling $D\to A$ moves downward and to the left.
Let
$$
r=\frac{V_1}{V_2}>1.
$$
The two adiabatic relations give
$$
T_B=T_A r^{\gamma-1},
\qquad
T_C=T_D r^{\gamma-1}.
$$
Because heat is exchanged at constant volume,
$$
Q_1=C_V(T_C-T_B),
\qquad
Q_2=C_V(T_D-T_A),
$$
where $Q_2$ denotes the positive amount rejected. Hence
$$
\begin{aligned}
\frac{Q_2}{Q_1}
&=\frac{T_C/r^{\gamma-1}-T_A}
{T_C-T_A r^{\gamma-1}}\\
&=\frac1{r^{\gamma-1}}.
\end{aligned}
$$
The <thermal efficiency> is therefore
$$
\boxed{
\eta=\frac W{Q_1}
=1-\frac{Q_2}{Q_1}
=1-\frac1{r^{\gamma-1}}}.
$$
For a fixed gas, $\gamma$ is fixed and the idealized efficiency increases monotonically with the compression ratio $r$, so it is maximized by making $r$ as large as the physical constraints allow.
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