The range plotted here is the straight-line distance $r$ from the launch point to the projectile. With launch speed $u$ and elevation $\theta$, the position at time $t$ is:

so, by Pythagoras, using $\cos^2\theta + \sin^2\theta = 1$:

Since $r$ is never negative, it has its turning points at the same times as $r^2$, which is much easier to differentiate. Setting the derivative to zero:

The root $t = 0$ is the launch itself. Dividing it out leaves a quadratic in $t$:

The range rises from zero, so the earlier root $t_-$ is a local maximum and the later root $t_+$ is the local minimum. Substituting $t_+$ back into the expression for $r^2$ gives the range at the minimum.

These roots are only real when the term under the square root is not negative:

Below this angle the range only ever increases, which is why the 40°, 50° and 60° curves have no minimum. At exactly 70.5° the two roots coincide and the curve has a stationary point of inflection. Above it, the projectile is thrown steeply enough that it comes back towards the launch point for a while before moving away again, so the 75°, 80° and 85° curves each dip. These points are marked with crosses on both graphs.

There is also a geometric way to see this. The derivative of $r^2$ is twice the dot product of the position and velocity vectors:

so the maximum and minimum occur at the two moments when the projectile is moving at right angles to the line joining it to the launch point.