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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).

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Xander Henderson
    Aug 21, 2021 at 18:54
  • $\begingroup$ ►$x+\sqrt{4-x^2}\ge x$ then there is a maximum above the diagonal between $(-2,-2)$ and $(2,2)$ ►Solving $y=x+\sqrt{4-x^2}$ you have $2x^2-2yx+y^2-4=0$ so $x$ have two distinct values $x_1,x_2$. ►Make the difference $x_1-x_2=0$ you get $-y^2+8=0$ which give you the maximum $y=2\sqrt2$ $\endgroup$
    – Piquito
    Aug 24, 2021 at 20:10

9 Answers 9

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Let $g(x) = x - \sqrt{4-x^2}$. Then, $f^2 + g^2 = 8$

And, since it is possible for $g$ to be zero, $\max(f)=\sqrt8=2\sqrt2$

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    $\begingroup$ I really like this one and the geometric proof shown above. Only thing about this one is that it seems like IVT, which is barely calculus, is kinda needed to assume g is zero. Otherwise, really cool solution! $\endgroup$
    – Copywright
    Aug 21, 2021 at 18:26
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    $\begingroup$ @Copywright One can intuitively see that $g$ can be zero from geometry (line $y=x$ and cemicircle $y=\sqrt{4-x^2}$ intersect somewhere in the first quadrant). More rigorously, you can just show that $x=\sqrt{4-x^2}$ has a positive solution ($x=\sqrt{2}$). $\endgroup$
    – Misha
    Aug 21, 2021 at 20:18
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I've sketched a purely geometric proof (tough it does informally use the rate of change). enter image description here

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By the RMS-AM means inequality the following holds, with equality iff $\,x=\sqrt{4-x^2}\,$:

$$ \require{cancel} \frac{x+\sqrt{4-x^2}}{2} \le \sqrt{\frac{\left(x\right)^2+\left(\sqrt{4-x^2}\right)^2}{2}} = \sqrt{\frac{\cancel{x^2} + 4 - \cancel{x^2}}{2}} = \sqrt{2} $$

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

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


$\text {Method}-\text{1b}$ \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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  • $\begingroup$ From $|u|\leq2\sqrt2$ you cannot conclude $\max\{u\}=2\sqrt2,$ only $\max\{u\}\leq2\sqrt2.$ To conclude equality you also need to show that there is some $x$ that makes $u=2\sqrt2.$ $\endgroup$
    – md2perpe
    Aug 21, 2021 at 22:05
  • $\begingroup$ @md2perpe let me write about the first method. Note that, $\Delta≥0$. This implies, $u=2\sqrt 2$ is possible. Then note note that, $(u-x)^2≥0$, this means $4-x^2≥0$ is also valid. $\endgroup$ Aug 21, 2021 at 22:36
  • $\begingroup$ What I mean is that if one only looks a the last implication, then it looks like you are concluding $\max\{u\}=2\sqrt2$ from only $|u|\leq2\sqrt2.$ That step is not correct in isolation. But together with some argument that this value can be reached (e.g. from theory of quadratics), the conclusion applies. $\endgroup$
    – md2perpe
    Aug 21, 2021 at 22:41
  • $\begingroup$ When there is a chain of implications, then one usually expects each implication to be valid in isolation. When a conclusion is made that does not only use the last statement, then it's best to use text to explain why that conclusion can be made. $\endgroup$
    – md2perpe
    Aug 21, 2021 at 22:59
  • $\begingroup$ I won't argue more about this. We know that the value $2\sqrt2$ can be reached by taking $x=\sqrt2.$ Then it's enough to show $u\leq2\sqrt2$ to conclude $\max u=2\sqrt2.$ $\endgroup$
    – md2perpe
    Aug 21, 2021 at 23:00
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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} $

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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 $x=\sqrt{2}$).

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  • $\begingroup$ (+1) This is very close to the approach I took. With the restriction of the domain and the explicit trigonometric formulas, this is a great answer (avoiding calculus). $\endgroup$
    – robjohn
    Aug 24, 2021 at 20:34
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Hopefully, this answer I added can be useful as well.

Let $a=x,~b=\sqrt{4-x^2}$, then we have

$$\begin{align}&(a+b)^2≤2(a^2+b^2)=8\\ \implies &\max \left\{a+b\right\}=2\sqrt 2.\end{align}$$

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  • $\begingroup$ ... where the first inequality is valid since $2(a^2+b^2)-(a+b)^2 = (a-b)^2 \geq 0.$ $\endgroup$
    – md2perpe
    Aug 21, 2021 at 22:12
  • $\begingroup$ @md2perpe $(a-b)^2≥0$ is always true, then where is the problem? $\endgroup$ Aug 21, 2021 at 22:22
  • $\begingroup$ There's no problem with that. I wrote "since", not "if". I couldn't see directly why $(a+b)^2 \leq 2(a^2+b^2)$ was valid, but after expanding $2(a^2+b^2)-(a+b)^2$ and getting $(a-b)^2$ which of course always $\geq 0,$ it was clear. So I added my comment so that others could get a clue. $\endgroup$
    – md2perpe
    Aug 21, 2021 at 22:31
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    $\begingroup$ @md2perpe You can also recognize it as $\frac{a+b}{2} \le \sqrt{\frac{a^2+b^2}{2}}$ which is the RMS-AM means inequality. $\endgroup$
    – dxiv
    Aug 22, 2021 at 1:39
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    $\begingroup$ @lonestudent No reason to delete. It is not much closer to RMS-AM than it is to AM-GM (if you reduce it to $\,2ab \le a^2+b^2\,$), and both can be derived from C-S for that matter. But it always helps to recognize that there is more than one way to frame it. $\endgroup$
    – dxiv
    Aug 31, 2021 at 6:45
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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 $2\sqrt{2}$.

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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:

  1. $m$, an arbitrary line parallel to and distinct from $\ell$;
  2. $N$, the polar point of $m$;
  3. $a$ and $b$, the parallel tangent lines to the intersection points of $\ell$ with $\gamma$ ($a$ and $b$ can be constructed by Pascal's theorem);
  4. $n$, the unique line through $N$ parallel to $a$ and $b$; and
  5. $P$ and $Q$, the intersection points of $n$ with $\gamma$.

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

Diagram of the construction of points whose tangents are parallel to a given diameter of a conic.

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.

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