Maxima and minima — using calculus — for IITJEE Advanced Mathematics

Problem:

The length of the edge of the cube $ABCDA_{1}B_{1}C_{1}D_{1}$ is 1. Two points M and N move along the line segments AB and $A_{1}D_{1}$ , respectively, in such a way that at any time t $(0 \leq t < \infty)$ we have $BM = |\sin{t}|$ and $D_{1}N=|\sin(\sqrt{2}t)|$ . Show that MN has no maximum.

Proof:

Clearly, $MN \geq MA_{1} \geq AA_{1} =1$ . If $MN=1$ for some t, then $M=A$ and $N=A_{1}$ , which is equivalent to $|\sin{t}|=1$ and $|\sin{\sqrt{2}t}|=1$ . Consequently, $t=\frac{\pi}{2}+k\pi$ and $\sqrt{2}t=\frac{\pi}{2}+n\pi$ for some integers k and n, which implies $\sqrt{2}=\frac{2n+1}{2k+1}$ , a contradiction since $\sqrt{2}$ is irrational. This is why $MN >1$ for any t.

We will now show that MN can be made arbitrarily close to 1. For any integer k, set $t_{k}=\frac{\pi}{2}+k\pi$ . Then, $|\sin{t_{k}}|=1$ , so at any time $t_{k}$ , the point M is at A. To show that N can be arbitrarily close to $A_{1}$ at times $t_{k}$ , it is enough to show that $|\sin(\sqrt{2}t_{k})|$ can be arbitrarily close to 1 for appropriate choices of k.

We are now going to use Kronecker’s theorem:

If $\alpha$ is an irrational number, then the set of numbers of the form $ma+n$ , where m is a positive integer while n is an arbitrary integer, is dense in the set of all real numbers. The latter means that every non-empty open interval (regardless of how small it is) contains a number of the form $ma+n$ .

Since $\sqrt{2}$ is irrational, we can use Kronecker’s theorem with $\alpha=\sqrt{2}$ . Then, for $x=\frac{1-\sqrt{2}}{2}$ , and any $\delta>0$ , there exist integer $k \geq 1$ and $n_{k}$ such that $k\sqrt{2}-n_{k} \in (x-\delta, x+\delta)$ . That is, for $\in_{k}=\sqrt{2}k+\frac{\sqrt{2}}{2}-\frac{1}{2}-n_{k}$ we have $|\in_{k}|<\delta$ . Since $\sqrt{2}(k+\frac{1}{2})=\frac{1}{2}+n_{k}+\in_{k}$ , we have

$|\sin(\sqrt{2}t_{k})|=|\sin \pi \sqrt{2}(k+\frac{1}{2})|=|\sin(\frac{\pi}{2}+n_{k}\pi+\in_{k}\pi)|=|\cos{(\pi \in_{k})}|$ .

It remains to note that $|\cos{(\delta\pi)}|$ tends to 1 as $\delta$ tends to 0.

Hence, MN can be made arbitrarily close to 1.

Ref: Geometric Problems on Maxima and Minima by Titu Andreescu, Oleg Mushkarov, and Luchezar Stoyanov.

Thanks to Prof Andreescu, et al !

Nalin Pithwa

This entry was written by nkpithwa, posted on May 17, 2016 at 9:29 am, filed under calculus, IITJEE Advanced Mathematics. Bookmark the permalink. Follow any comments here with the RSS feed for this post. Post a comment or leave a trackback: Trackback URL.

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