At some geocentric latitude $\phi$ and longitude $\lambda$, the elevation or altitude $a$ of a star with right ascension $\alpha$ and declination $\delta$ is given by
$$\begin{aligned}
\sin a = \sin\phi \sin \delta + \cos \phi \cos\delta \cos h&&&&&&(1)
\end{aligned}$$
where
$$\begin{aligned}
h=\theta_\phi - \alpha\qquad&&\qquad&&\qquad&&\quad\,&&(2)
\end{aligned}$$
is the hour angle to the star, expressed in terms of the local sidereal time $\theta_{\phi}$ and the right ascension $\alpha$.
Differentiating (1) with respect to time yields
$$\begin{aligned}
\cos a \frac{da}{dt} = &
\phantom{+}\,(\cos\phi\sin\delta - \sin\phi \cos\delta \cos h)\,\frac{d\phi}{dt} \\
& + \,(\sin\phi \cos\delta - \cos\phi \sin\delta \cos h)\,\frac{d\delta}{dt} \\
& - \cos\phi \cos\delta \sin h\,\frac{dh}{dt} &\!\!\!(3)
\end{aligned}
$$
While not quite zero, the time derivatives of latitude $\phi$ and declination $\delta$ are negligibly small compared to the $2\pi$ radians per sidereal day time derivative of hour angle $h$. For all practical purposes, the above thus reduces to
$$\begin{aligned}
\cos a \frac{da}{dt} = - \cos\phi \cos\delta \sin h\,\frac{dh}{dt} \quad\!&&&&&&(4)
\end{aligned}
$$
The left hand side of (4) is zero at extrema of elevation angle. Thus we're looking for conditions that make the right hand side of (4) equal zero. Since $dh/dt$ is nonzero, the right hand side is zero only if one or more of $\cos \phi$, $\cos \delta$, or $\sin h$ is zero. The first two cases ($\cos \phi = 0$ and $\cos \delta = 0$) represent conditions where elevation is constant. The only condition of interest is $\sin h = 0$, which means an hour angle of 0° or 180° (hour angle is constrained to lie between 0° (inclusive) and 360° (exclusive)).
Since latitude $\phi$ and declination $\delta$ both lie between -90° and +90°, the condition $h=180^{\circ}$ represents the minimum possible elevation, while $h=0$ represents the maximum possible elevation. From (2), $h=0$ means $\theta=\alpha$, or local sidereal time being equal to right ascension.
