# Optimisation problem choose x to minimize y

I have stumbled upon a sample maths question during my revision, and I have no idea how to solve it. Can anyone help or guide me along?

Given a piece of rectangular paper of 11 cm by 8.5 cm. The lower left-hand corner is to be folded over to reach the top edge as shown in the diagram. How would you fold it so as to minimize the length of the crease y ? In other words, how would you choose x to minimize y ? (Try it with paper yourself.)

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It's almost obvious that x does not depend on the width of the strip, and actually ${x \over A}$ where A is rectangle's height (here 8.5) must be a constant. With the help of a CAS, and writing equations from conditions from the drawing (Pythagoras theorem, etc.) one finds this constant is ${3 \over 4}$. Now, I'm just looking for a geometric proof :-) – Jean-Claude Arbaut Mar 14 '13 at 19:46

To make life easier, let the width of the paper be $1$; we can scale up to $8.5$ at the end. Look at the "missing" triangle at the bottom left. Let the angle on the right of that triangle, the angle of the fold, be $\theta$.

Then what you have called $x$ is $y\sin\theta$.

Now look at the little triangle at top left. Its bottom angle, by angle-chasing, is $2\theta$. Its hypotenuse is $y\sin\theta$, and the "adjacent" side is $1-y\sin\theta$. So we obtain $$\frac{1-y\sin\theta}{y\sin\theta}=\cos 2\theta.$$ Solve for $y$. We get $$y=\frac{1}{\sin\theta(\cos 2\theta +1)}.$$ The identity $\cos 2\theta=2\cos^2\theta -1$ improves this to $$y=\frac{1}{2\sin\theta\cos^2\theta}$$ We want to minimize $y$. So we want to maximize $\sin\theta\cos^2\theta$, that is, $2\sin\theta(1-\sin^2\theta)$. So we are in essense minimizing $2t-2t^3$.

Differentiate as usual. We find that for smallest $y$, we need $\sin\theta=\frac{1}{\sqrt{3}}$. The corresponding $y$ is $\frac{3\sqrt{3}}{4}$, which (almost) gives $\frac{3}{4}$ for your $x$, except that we need to scale up by the factor $8.5$. The prettiness of the number $\frac{3}{4}$ undoubtedly means that the calculus way, though simple enough, is not optimal.

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Almost isn't good enough. Calculus wins. – PyRulez Mar 14 '13 at 21:08
This $3 \over 4$ is frustrating, isn't it ? :-) – Jean-Claude Arbaut Mar 14 '13 at 21:42
Yes, a bit tantalizing, though the calculus way was pretty easy. – André Nicolas Mar 14 '13 at 21:48

The length of the paper is a red herring, you can crease even an infinitely long strip of paper. Considering the infinitely long strip, and moving the point where the crease hits the bottom margin to the right, you see the maximal $y$ is just $\infty$. Moving that point to the left diminishes $y$, which gets minimal when $x$ hits the top, i.e., $x = 8.5$ with $y = 8.5 \sqrt{2}$.

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But we are to minimize the length of the crease, not $y$, and we can do better. – Ross Millikan Mar 15 '13 at 14:07