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).
 A: By the RMS-AM means inequality the following holds, with equality iff $\,x=\sqrt{4-x^2}\,$:
$$
\require{cancel}
\frac{x+\sqrt{4-x^2}}{2} \le \sqrt{\frac{\left(x\right)^2+\left(\sqrt{4-x^2}\right)^2}{2}} = \sqrt{\frac{\cancel{x^2} + 4 - \cancel{x^2}}{2}} = \sqrt{2}
$$
A: Let $x = 2 \cos \theta$
Then  $x + \sqrt{4 - x^2} = 2 \cos \theta + 2 \sin \theta = 2 (\cos \theta + \sin \theta)$
Now, the trigonometric function $ a \cos \theta + b \sin \theta $ satisfies:
$$-\sqrt{ a^2 + b^2 } \le a \cos \theta + b \sin \theta \le \sqrt{a^2 + b^2} $$
Therefore, $\cos \theta + \sin \theta \le \sqrt{2} $
Thus $x + \sqrt{4 - x^2} \le 2 \sqrt{2} $
A: Let $g(x) = x - \sqrt{4-x^2}$.  Then, $f^2 + g^2 = 8$
And, since it is possible for $g$ to be zero, $\max(f)=\sqrt8=2\sqrt2$
A: Hopefully, this answer I added can be useful as well.
Let $a=x,~b=\sqrt{4-x^2}$, then we have
$$\begin{align}&(a+b)^2≤2(a^2+b^2)=8\\
\implies &\max \left\{a+b\right\}=2\sqrt 2.\end{align}$$
A: I've sketched a purely geometric proof (tough it does informally use the rate of change).

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

$\text {Method}-\text{1b}$
\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}

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

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

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

$\text {Method}-4$
$$\begin{align}&u^2-4=2\sqrt{x^2(4-x^2)}, \thinspace x≥0\\ 
\implies &x^2+(4-x^2)≥2\sqrt{x^2(4-x^2)}\\
\implies &u^2-4≤4\\
\implies &\max \left\{u\right\}=2\sqrt 2.
\end{align}$$
A: You can transform the variable $x$ to something that makes it more clear. Let $x=2\sin t$. (Since $x$ can be in $[-2,2]$, restrict $t$ to $[-\pi/2,\pi/2]$.) Then:
$$
\begin{align}
f(t)&=2\sin t+2\cos t\\
&=2\sqrt{2}\left[\sin(t)\cos(\pi/4)+\cos(t)\sin(\pi/4)\right]\\
&=2\sqrt{2}\sin(t+\pi/4)\\
&\leq2\sqrt{2}
\end{align}$$
And equality happens when $t=\pi/4$ (when $x=\sqrt{2}$).
A: Use the Cauchy-Schwarz inequality, $(a^2+b^2)(c^2+d^2)\ge (ac+bd)^2$. Let $a=b=1$ and $c=x$ and $d=\sqrt{4-x^2}$. You can find that the maximum is $2\sqrt{2}$.
A: The graph of the function comprises half an ellipse. The $x$-axis is a diameter for this ellipse. Finding the maximum of the function entails finding tangents to the ellipse parallel to the $x$-axis.
So suppose given a conic section $\gamma$ with diameter $\ell$. Construct, with straightedge and parallel rule:

*

*$m$, an arbitrary line parallel to and distinct from $\ell$;

*$N$, the polar point of $m$;

*$a$ and $b$, the parallel tangent lines to the intersection points of $\ell$ with $\gamma$ ($a$ and $b$ can be constructed by Pascal's theorem);

*$n$, the unique line through $N$ parallel to $a$ and $b$; and

*$P$ and $Q$, the intersection points of $n$ with $\gamma$.

Then $P$ and $Q$ are the two points on $\gamma$ whose tangents are parallel to $\ell$.

The point of this exercise in synthetic geometry is to show that finding the maximum of the function reduces to a question of affine geometry, so that we are free to make affine-linear transformations of the coordinates to simplify to, e.g., the case of the circle. I think that's the spirit of what several other of the answers have done.
