Here is a question on a test I asked recently in a Calculus I class I teach on Saturdays. These are students who are planning to major in Math, and they are already taking AP Calculus BC in their high schools, but some want to take it a step further and practice more (eg: go deeper) on the weekends. Anyway, here is the question and, with permission of the student, I have posted his solution. I have my own solution too, but it is different in scope, so I do not want to post it yet. Any other approaches to this problem? As a hint, I will say that my solution uses the Weierstrass Theorem (the extreme value theorem). I tried to add more tags to the post, but apparently more points are required.

A sign, posted on a wall, has its upper and lower edges represented by heights m and n, respectively, with reference to a reader's vision. How far from the wall should a person reading the sign be located so that the visual angle determined by the pupil and the edges is at a maximum?

A student solved it as follows:

Let us suppose $m>n$ (in other words, m represents the height of the top edge and r represents the height of the bottom edge, so $m>n$.

Thus, $$\theta_1=\arctan\left(\frac{m}{d}\right), \theta_2=\arctan\left(\frac{n}{d}\right)$$

We are trying to maximize the difference between $\theta_1$ and $\theta_2$. Essentially we need to maximize $\arctan\left(\frac{m}{d}\right)-\arctan\left(\frac{n}{d}\right)$. We take the derivative with respect to $d$ to get: $$\frac{n}{n^2+d^2}-\frac{m}{m^2+d^2}$$

Set this equal to $0$ and solve for $d$ to get: $$d=\sqrt{mn}$$

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    $\begingroup$ This looks like a problem for MSE rather than MESE. $\endgroup$
    – Sue VanHattum
    Oct 26 '21 at 22:54
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    $\begingroup$ Well if $m>0>n$, so the sign extends from above my eye to below it, then I'd put my eye right up against the sign and get an angle of $\pi$. So a solution should pay attention to the signs of $m$ and $n$ $\endgroup$ Oct 27 '21 at 1:04
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    $\begingroup$ This is a “famous” optimization problem and can be solved with or without calculus. See en.m.wikipedia.org/wiki/… $\endgroup$
    – KCd
    Nov 26 '21 at 8:49

I always find it fun to try to avoid calculus in problems like this, so if you want to show your students some applications of geometry and trig, how about something like this: use geometric reasoning to see that the angle is maximized when your eye is at the bottom of the circle containing your eye and the poster (since, looking at the angle at the center of the circle, you want $mn$ to be a chord of the smallest such circle).

Now we get $\tan\phi=n/d$, $\tan\theta=\frac{m-n}{2d}$, and $\tan(\theta+\phi)=m/d$. The angle sum formula and algebra take care of the rest. (I tried to resize the picture but sticking on an "s" was too small and an "m" previewed as almost all black! If I figure it out I'll edit.)

enter image description here

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    $\begingroup$ Any reason for the downvote? $\endgroup$
    – Thierry
    Nov 1 '21 at 13:51

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