<s>[INST] Determine the maximum of $f(x) = x + \sqrt{4-x^2}$ without calculus I was given this problem in a Calc BC course while we were still doing review, so using derivatives or any sort of calculus was generally forbidden. We were only doing review because a lot of people hadn't taken Calc AB due to scheduling issues so the teacher felt he should at least cover some of what would be covered in AB for a while.
Of course using calculus this problem is quite trivial. $f'(x) = x/\sqrt{4-x^2}$, set this equal to zero and the maximum occurs at ($\sqrt2$, $2\sqrt2$). But again, this method was not allowed.
I immediately recognized the $\sqrt{4-x^2}$ as a semicircle with maximum value at 2, so I knew that because x was being added, the value must be greater than 2 at the maximum. Also, the x value must be less than two.
I then attempted to rearrange the equation such that it could be written in polar coordinates but, letting $y=f(x)$ caused $x^2$ to cancel out and I was left with $y^2 = 2x\sqrt{4-x^2}$ which only seems to complicate the equation further.
I've been stuck at this point for a while as I haven't thought too much about the problem, but my teacher dismissed the problem as he didn't realize he had assigned it. Regardless, this was in the pre-calculus portion of our textbook, so I assume that there has to be a way to solve it (our textbook only gives solutions for odd problems).
 [/INST] $\text {Method}-\text{1a}$
$$\begin{align}&u=x+\sqrt {4-x^2},\thinspace u-x≥0\\ \implies &(u-x)^2=4-x^2\\ \implies &u^2-2ux+2x^2-4=0\\ \implies &2x^2-2ux+(u^2-4)=0\\ \implies &\Delta=u^2-2(u^2-4)≥0\\
\implies &u^2≤8\\
\implies &|u|≤2\sqrt 2\\ \implies &\max \left\{u\right\}=2\sqrt 2. \end{align}$$

$\text {Method}-\text{1b}$
\begin{align}&u=x+\sqrt {4-x^2},\thinspace u-x≥0\\ \implies &(u-x)^2=4-x^2\\ \implies &u^2-2ux+2x^2-4=0\\ 
\implies &2x^2-2ux+(u^2-4)=0\\ 
\implies &2\left(x-\frac u2\right)^2+\frac{u^2-8}{2}=0\\
\implies &u^2-8≤0\\
\implies &\max \left\{u\right\}=2\sqrt 2. \end{align}

$\text {Method}-\text{2a}$
$$\begin{align}&u^2=4+2x\sqrt {4-x^2}\\
\implies &x\sqrt{4-x^2}=a, \thinspace a=\frac{u^2-4}{2}\\
\implies &\frac{a^2}{x^2}+x^2=4, \thinspace x≠0 \\
\implies &\left(\frac ax-x\right)^2+2a=4 \\
\implies &\left(\frac ax-x\right)^2=4-2a≥0\\
\implies &a=\frac {u^2-4}{2}≤2\\ \implies &u^2≤8\\ \implies &\max \left\{u\right\}=2\sqrt 2.
\end{align}$$

$\text {Method}-\text{2b}$
$$\begin{align}&u^2=4+2x\sqrt {4-x^2},\thinspace x≠0\\
\implies &x\sqrt{4-x^2}=a, \thinspace a=\frac{u^2-4}{2}\\
\implies &\frac{a^2}{x^2}+x^2=4≥2\sqrt{\frac{a^2}{x^2}\times x^2}=2a \\
\implies &a=\frac {u^2-4}{2}≤2\\
\implies &\max \left\{u\right\}=2\sqrt 2.
\end{align}$$

$\text {Method}-3$
$$\begin{align}&\frac {u^2-4}{2}=\sqrt{x^2(4-x^2)},\thinspace x≥0\\
\implies &\frac {u^2-4}{2}=\sqrt{4x^2-x^4}\\
\implies &\sqrt{4-(x^2-2)^2}≤2\\
\implies &\frac {u^2-4}{2}≤2\\
\implies &\max \left\{u\right\}=2\sqrt 2.
\end{align}$$

$\text {Method}-4$
$$\begin{align}&u^2-4=2\sqrt{x^2(4-x^2)}, \thinspace x≥0\\ 
\implies &x^2+(4-x^2)≥2\sqrt{x^2(4-x^2)}\\
\implies &u^2-4≤4\\
\implies &\max \left\{u\right\}=2\sqrt 2.
\end{align}$$
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