# 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).

• Comments are not for extended discussion; this conversation has been moved to chat. Commented Aug 21, 2021 at 18:54
• ►$x+\sqrt{4-x^2}\ge x$ 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 $y=2\sqrt2$ Commented Aug 24, 2021 at 20:10

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

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

• 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! Commented Aug 21, 2021 at 18:26
• @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 ($x=\sqrt{2}$). Commented Aug 21, 2021 at 20:18

I've sketched a purely geometric proof (tough it does informally use the rate of change).

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

\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}

• 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 $u=2\sqrt2.$ Commented Aug 21, 2021 at 22:05
• @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. Commented Aug 21, 2021 at 22:36
• 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. Commented Aug 21, 2021 at 22:41
• 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. Commented Aug 21, 2021 at 22:59
• 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 $\max u=2\sqrt2.$ Commented Aug 21, 2021 at 23:00

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

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

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 $$x + \sqrt{4 - x^2} \le 2 \sqrt{2}$$

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 $$x=\sqrt{2}$$).

• (+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).
– robjohn
Commented Aug 24, 2021 at 20:34

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}

• ... where the first inequality is valid since $2(a^2+b^2)-(a+b)^2 = (a-b)^2 \geq 0.$ Commented Aug 21, 2021 at 22:12
• @md2perpe $(a-b)^2≥0$ is always true, then where is the problem? Commented Aug 21, 2021 at 22:22
• 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. Commented Aug 21, 2021 at 22:31
• @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.
– dxiv
Commented Aug 22, 2021 at 1:39
• @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.
– dxiv
Commented Aug 31, 2021 at 6:45

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 $$2\sqrt{2}$$.

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 $$\ell$$.

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.