calculusderivativesgeometryreal-analysis
One corner of a rectangular sheet of paper is folded over so as to reach the opposite edge(lengthwise) of the sheet. If area of the folded paper is minimum, show the crease divided the width in 2:3.
Note: We can use double derivative concept to encounter this problem. I don't understand the geometry the sum portrays. Please if possible post an image.
Just a hint about the set-up:
If say the top left corner is at $A=(0,11)$ and it gets folded down to the right side $x=8.5$ so it winds up at $P(t)=(8.5,t),$ then the crease would be the intersection of the perpendicular bisector of segment $A\ P(t)$ with the sheet of paper. So parametrize this situation and hopefully proceed to get an expression for the length $L(t)$ of that intersection, so as to apply usual minimization techniques.
My suggestion is to compute AD' and AB' as a function of AD and $\alpha$. $\Delta AOD' \sim \Delta ADA' \to AD' \times AD = AO\times AA'$
and note that $AA' = \frac{AD}{\cos \alpha}$ and $AO = \frac{1}{2}AA'$
$\therefore AD' = \frac{AA'^2}{2AD}=\frac{AD}{2\cos^2 \alpha}$
$AB' = \frac{AD'}{\tan \alpha} = \frac{AD}{2\cos^2 \alpha} \times \frac{\cos \alpha}{\sin \alpha}=\frac{AD}{\sin 2\alpha}$
$S_{AB'D'} = \frac{1}{2}AB' \times AD' = \frac{AD^2}{2\cos \alpha \sin 2\alpha}$
Since AD is constant, maximizing $S_{AB'D'}$ is to maximize the expression of $\alpha$. You can find the bound of $\alpha$ by moving (1) D' to D and (2) B' to B combining with the assumption 3:7 given. Then, the min or max could be easily obtained.
Best Answer
Have you picked up a piece of paper and folded it? The dotted line shows where the triangle came from. The triangles above and below the highest diagonal line are supposed to be congruent. You want to maximize the area of the folded part, which minimizes the area left.