A standard $8.5$ inches by $11$ inches piece of paper is folded so that one corner touches the opposite long side and the crease runs from the adjacent short side to the other long side, as shown in the picture below. What is the minimum length of the crease?
So I know this problem involves a lot of Calculus and Optimization, but I'm really lost to be completely honest. Any help or hints would be super helpful! Thanks!
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Since $a^{2}+b^{2}=crease^{2}$ we can minimize $a^{2}+b^{2}$ rather than its square root. Using the Lagrange multiplier method, $a^{2}+b^{2}$ will be the expression. From the lower right triangle, we can see that $c^{2}+(8.5-b)^{2}=b^{2}.$ That equation can be a constraint, but we added a variable (variable c) and we must have a constraint for every variable after the first one. So we need another constraint. Looking at the left edge, we can see that there is a right triangle with $8.5$ as one edge (dotted green) and $11-(11-a)-c$ as the base. So a second constraint can be $(a-c)^{2}+8.5^{2}=a^{2}.$
$$\mathscr{L}(a,b,c,\lambda_{1},\lambda_{2})=a^{2}+b^{2}-\lambda_{1}(c^{2}+(8.5-b)^{2}-b^{2})-\lambda_{2}\left((a-c)^{2}+8.5^2-a^{2}\right)$$
$$\nabla\mathscr{L}=\left(\begin{array}{c} 2\;c\;\lambda_{2}+2a\\ 2\;b+17\;\lambda_{1}\\ -2\;c\;\lambda_{1}+2\;\lambda_{2}\;\left(a-c\right)\\ b^2-c^2-\left(-b+\frac{17}{2}\right)^2\\ 2ac-c^2-\frac{289}{4} \end{array}\right)=\mathbf{0}$$
Solve these simultaneously and keep the positive values for $a$ and $b.$ $a=51\sqrt{2}/8$ and $b=51/8.$ Therefore, $$Crease= \sqrt{a^2+b^2}\approx 11.0418$$
As you can see in the figure above, $AC$ is the crease you are looking for. By superimposing, we observe that triangles $ACD $ and $ACB$ are congruent. Therefore let $\angle DAC = \angle BAC = \alpha$.
Let $AB = y , BC = x$
Then we have $x = y \tan \alpha\quad$(see triangle $ABC$)
and $ \quad y \sin 2\alpha = EB = 8.5$
Note that $y$ will vary continuously from $11$ inches to the lower limit of $y = 8.5.$
We want to minimise crease $AC$, i.e. to minimise $x^2 + y^2 = y^2(\tan^2 \alpha + 1) = \big(\frac{8.5}{\sin 2 \alpha}\big)^2\big( \tan^2 \alpha + 1 \big)$, which is a function of $\alpha$.
I hope you can proceed now using calculus to find minima.
You have to minimise $\dfrac{1}{\sin^2 2 \alpha \cos^2 \alpha} =\bigg({\dfrac{2}{\cos 2 \alpha + 1}}\bigg)\bigg(\dfrac{1}{1-\cos^2 2 \alpha}\bigg) = \bigg(\dfrac{2}{(1-t^2)(1+t)}\bigg) = \bigg(\dfrac{2}{(1-t)(1+t)^2}\bigg) ... (1)$,
where $t = \cos 2 \alpha$.
Note that $t$ can't be $1$ as $\alpha \neq 0 $, so expression is well defined.
I leave it as an exercise to prove that this has minima at $t = \frac{1}{3}$, and minima of above expression $(1)$ is $\approx 1.687$.
Thus the final answer is $\big(8.5^2 \times 1.687\big)^{0.5} \approx 11.04 $ inches.
Here is a graph of the function to verify the minima.