Maximization of velocity involving trigonometric functions I have a physics question but my main doubt is regarding the math part of it, namely the maximization of the velocity. I'll attach a diagram of the question here for reference.

I am given the values of coefficients of friction, $\mu_s$(static friction coefficient) = $\frac{1}{\sqrt3}$, $\mu_k$(kinetic friction coefficient) = $\frac{1}{\sqrt5}$
It is given that the point A describes a circle in the vertical plane at a constant speed $v_A$, the question asks me to find the maximum speed $v_A$ such that the block B doesn't slide on the platform.
Without showing, the main physics part of it, (I can provide the working if needed though), what I'm obtaining from this is $$v_A^2=\frac{4}{\sqrt3 \cos \theta +\sin \theta}$$
Now, to maximize $v_A$, clearly I need to minimize the denominator here. Now, I know, $$-2 \leq\sqrt3 \cos \theta+\sin \theta \leq 2$$
Now, $v_A^2$ has to be positive and so $-2$ clearly cannot be the required minimum value of the denominator. However, I find that the answer to this question is that $v_A = \sqrt2$ which means that $v_A^2=2$ and that means that the value of $\sqrt3 \cos \theta + \sin \theta$ has been taken to be $2$ which is the maximum value of it. But, shouldn't that make the value of $v_A$ minimum, as the denominator becomes maximum?
EDIT: According to http://ilin.asee.org/Conference2007program/Papers/Conference%20Papers/Session%202B/Liang.pdf
"The normal analytical approach to this problem has two steps. In the first step, using the x- and the y- components (or the normal and the tangential components) of the equation of motion, an equation of maximum velocity vA without sliding is found as
$$(v_{A(max)})^2 =\frac{\mu_s g \rho} {(\cos \theta + \mu_s \sin \theta)}$$
where $g$ is the gravitational acceleration, $\rho$ is the radius of curvature, θ is the angle of crank OA as shown in the Fig 6, and max VA is the velocity permissible by the friction for the particular crank angle θ. In the second step, the minimum of $(v_{A(max)})^2$ and the corresponding θ are found using calculus or Excel Solver. Here the above two sentences must be written carefully to tell what is maximized and what is minimized. The same possible confusion can arise in the students."
 A: $(v_{A(max)})^2 =\frac{\mu_s g \rho} {(\cos \theta + \mu_s \sin \theta)}$ means that $(v_{A})^2 \leq \frac{\mu_s g \rho} {(\cos \theta + \mu_s \sin \theta)}$. Since this should hold for all possible values of $\theta$, we have
$$(v_{A})^2 \leq \min{\left(\frac{\mu_s g \rho} {(\cos \theta + \mu_s \sin \theta)}\right)}$$
Thus the maximum value of $v_A^2$ is actually the minimum of $\left(\frac{\mu_s g \rho} {(\cos \theta + \mu_s \sin \theta)}\right)$ which is nothing but $(v_{A(max)})^2$.
Thus you need to minimise $(v_{A(max)})^2$ to get the maximum value of $(v_{A})^2$.
The minimum value you got for $\frac{4}{\sqrt3 \cos \theta +\sin \theta}$ was right and as such $v_A (max)=\sqrt{2}$
A: In order for the block not to slide, the force of static friction must match the horizontal component of the centripetal force towards the point of rotation. When you solved for the maximum speed that achieves this, you got an expression in terms of the angle of rotation, $\theta$:
$$v_A=\sqrt{\frac{4}{\sqrt3 \cos \theta +\sin \theta}}$$
Now, the speed $v_A$ is constant, so in order for the block not to slide, $v_A$ needs to be such that the block would not slide at any $\theta$, so you need to take the minimum of $\sqrt{\frac{4}{\sqrt3 \cos \theta +\sin \theta}}$, which occurs at $\theta= \frac{\pi}{6}$ and $v_A=\sqrt{2}$.
In essence, the value for $v_A$ represents the maximum speed at a particular angle $\theta$ for which the block does not move, but you need to take a minimum of it so that the block never moves.
