# find the maximum of sum of roots without calculus

A question asks me to evaluate the maximum and minimum value of the following function:

$$y=\sqrt{x}+\sqrt{27+x}+\sqrt{13-x}, 0\le x\le13$$

I have tried to compute y using $$x=0,1,2,...,13$$ I believe such maximum occurs at x=9, and the minimum value lies on boundary points, but how can I prove that without calculus?

Thanks!

• When $x=6.1$ it's more than at $6.$ Looks more like max is at $x=9,$ where $y=11.$ – coffeemath Dec 6 '16 at 5:26
• Yeah you are right. – Ricky Choi Dec 6 '16 at 5:29
• YJust an observation: shifting $f$ by $9$ to the left we have $\sqrt{x+3^2}+\sqrt{6^2+x}+\sqrt{2^2-x}$ which seems to have its maximum in $x=0$. – Michael Hoppe Dec 6 '16 at 6:59
• Further note: it seems that $\sqrt{x+(n+1)^2}+\sqrt{x+(n(n+1))^2}+\sqrt{n^2-x}$ has a maximum in $x=0$. – Michael Hoppe Dec 6 '16 at 7:23
• Just one more: $n^2+(n+1)^2+(n(n+1))^2=(n^2+n+1)^2$. – Michael Hoppe Dec 6 '16 at 12:09

Here's a partial answer. Let $f(x)=\sqrt{x}+\sqrt{27+x}+\sqrt{13-x}$ for $0\leq x\leq 13$. Note that if $0\leq x< 13/2$, then $f(13-x)>f(x)$, since the first and third terms of $f(13-x)$ and $f(x)$ are the same (just swapped) and the second term of $f(13-x)$ is larger. So $f(x)$ reaches its minimum somewhere in $[0,13/2]$, and its maximum somewhere in $[13/2,13]$.

For the minimum, note that $$(\sqrt{x}+\sqrt{13-x})^2=13+2\sqrt{x(13-x)}=13+2\sqrt{169/4-(13/2-x)^2},$$ so $$f(x)=\sqrt{13+2\sqrt{169/4-(13/2-x)^2}}+\sqrt{27+x}.$$ This makes it clear that $f(x)$ is increasing on $[0,13/2]$ (since both terms are increasing), so the minimum is at $x=0$.

The maximum seems much harder to find, and I don't know how to find it without calculus...

Shifting the OP's function $9$ to the left we get a function of the form $$g(x)=\sqrt{x+(n+1)^2}+\sqrt{x+\bigl(n(n+1)\bigr)^2}+\sqrt{n^2-x}.$$

Now it's easy to prove without calculus simply by squaring that $\sqrt{x+a^2}\leq \frac{1}{2a}x+a$ and $\sqrt{a^2-x}\leq-\frac{1}{2a}x+a$ for positive $a$ and $-a^2< x<a^2$. Equality occurs iff $x=0$. From here

$$\sqrt{x+(n+1)^2}\leq\frac{1}{2(n+1)}x+n+1,$$ $$\sqrt{x+\bigl(n(n+1)\bigr)^2}\leq\frac{1}{2n(n+1)}x+n(n+1),$$ $$\sqrt{n^2-x}\leq-\frac{1}{2n}x+n.$$

Now adding up yields $g(x)\leq n+1+n(n+1)+n$ with equality iff $x=0$.

Hence the OP's function (with $n=2$) has its global maximum in $x=9$ of value $11$.

• I've edited right now. – Michael Hoppe Dec 11 '16 at 21:59
• Ah, that's very nice. – Eric Wofsey Dec 11 '16 at 22:23
• Thanks. So together with your answer the question is solved completely. – Michael Hoppe Dec 11 '16 at 22:37

I don't know if this is rigorous enough, but you can take the function and notice that when there is a maximum on the domain of $[0,13]$, because we have two increasing functions and one decreasing function all added together, our function only has one critical point (something we observe, not proved), not considering the endpoints. When we let $x_0$ be our maximum we notice that $$\frac{f(x)-f(x_0)}{x-x_0} =\begin{cases} >0 & \text{if }x<x_0 \\ <0 & \text{if }x>x_0 \end{cases}$$ Now obviously if $x = x_0$ then this is undefined, but if we choose any value on the left side of $9$, say $8.5$ we notice that we get a positive value, and for anything on the right side, like $9.5$ we end up with a negative value. This shows that $9$ is a local maximum.

I would say that the problem with this is that it hasn't verified that some other point isn't also a maximum, but I think you'll run into that problem as long as you aren't considering the derivative. So although it may not be perfect, it's a good start.

The usage of this is essentially looking at the slope of the line before and after some point, which you can argue is moving towards the realm of calculus, but it stays away from derivatives.

Edit: If we do indeed believe there is only the one critical point, you will find the minimum at the boundary point like you stated, so simply check the two of them to find which point is smaller.

• But how do you know that $\frac{f(x)-f(x_0)}{x-x_0}$ has the properties you state for $x_0=9$ and any $x$ sufficiently close to $9$? This seems rather nontrivial to prove. – Eric Wofsey Dec 6 '16 at 5:51
• I didn't say that it was perfect, only that it is true, noticing that there is only one critical point is key due to the fact, as mentioned above, that two pieces ($\sqrt{x}$ and $\sqrt{27 + x}$) are strictly increasing with the other piece strictly decreasing ($\sqrt{13-x}$), so you can be confident that there is only maximum. That maximum, when you find it, will have the property as shown. – Mitch Dec 6 '16 at 5:58
• The fact that two pieces are increasing and one piece is decreasing does not mean that there must be only one local maximum. You need to know something about the second derivatives as well. – Eric Wofsey Dec 6 '16 at 6:07
• If you were talking about three random functions, then yes that may be true, but it does hold in this case. I guess I was too aggressive in wanting to assert my answer and can admit that it wouldn't hold in a more general setting. – Mitch Dec 6 '16 at 6:14