Is it permissible to locate the abscissa of extreme points of $y=f(x)$ by powering the function first for the sake of simplicity? Let's take a simple example as follows,
$$R=\sqrt{A^2+B^2 +2AB\cos \theta}$$
It represents the magnitude relation of vectors $\vec A$, $\vec B$, and $\vec R$ which is $\vec A +\vec B$. And we have to find $\theta$ to make the longest and shortest $R$.
We can find $\theta$ from $\frac{\textrm{d}R}{\textrm{d}\theta}=0$ as a normal method. However, I think it will be simpler if we square the $R$ first and solve $\frac{\textrm{d}R^2}{\textrm{d}\theta}=0$.
For general case, is it permissible to do such a method?
 A: Let $q(x)$ be a non-negative function of $x$. Then $q(x)$ reaches a maximum (local or global) at $x=a$ if and only if $(q(x))^2$ reaches a maximum (local or global) at $x=a$. The same is true of we replace "maximum" everywhere by "minimum."
The same is true if instead of a function $q(x)$ of one variable, we are looking at a function $q(x_1,\dots,x_n)$ of several variables.
This is because of the fact that the function $f(t)=t^2$ is an increasing function on the interval $[0,\infty)$. So for any $x$ and $a$, we have $q(x)\le q(a)$ if and only if $(q(x))^2 \le (q(a))^2$.
The above remarks can be a useful simplifying device in max/min problems. 
Remark: More generally, if $n$ is an even positive integer, and $q(x)$ is a non-negative function, then $q(x)$ reaches a maximum at $x=a$ if and only if $(q(x))^n$ reaches a maximum at $x=a$.
If $n$ is an odd positive integer, the situation is even nicer. We get that without conditions on $q(x)$, this reaches a maximum at $x=a$ if and only if $(q(x))^n$ does.
More generally, if $G(t)$ is an increasing function, then $q(x)$ reaches a maximum at $x=a$ if and only if $G(q(x))$ does. 
