calculus
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 ends on the adjacent short side, as shown in the picture below. What is the minimum length of the crease?
This is, in theory, a simple optimisation problem, however, my approach quickly deteriorated into angle chasing. How should I approach the question?
Thanks!
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.
The volume may be calculated with Cavalieri's principle and a bit of care.
Place the cone with the center of its base at the origin. For $-r \leq x \leq r$, let $A(x)$ denote the area of the parabolic cross-section "at $x$".
Note that:
The area under the parabola is two-thirds the area of the circumscribing rectangle.
The width of the circumscribing rectangle is $2\sqrt{r^{2} - x^{2}}$.
A longitudinal section of the cone has sides $\sqrt{r^{2} + h^{2}}$ and base $2r$, so by similar triangles the circumscribing rectangle has slant height $(r - x)\sqrt{r^{2} + h^{2}}/(2r)$.
Consequently, $$ A(x) = \tfrac{2}{3} \cdot 2\sqrt{r^{2} - x^{2}} \cdot (r - x) \frac{\sqrt{r^{2} + h^{2}}}{2r} = \frac{2\sqrt{r^{2} + h^{2}}}{3r}\, \sqrt{r^{2} - x^{2}}\, (r - x). \tag{1} $$
Now for the bit of care: A thin planar slice meeting the cone's diameter at $x$ and $x + dx$ is tilted with respect to the $x$-axis, and by similar triangles has thickness $$ dv = \frac{h}{\sqrt{r^{2} + h^{2}}}\, dx \tag{2} $$ rather than $dx$. Multiplying (1) and (2), the volume of the planar slice of cone at location $x$, with $-r \leq x \leq r$, is $$ dV = A(x)\, dv = \frac{2h}{3r}\, \sqrt{r^{2} - x^{2}}\, (r - x)\, dx. \tag{3} $$ The volume of the cone is the integral $$ \text{Vol} = \int_{-r}^{r} \frac{2h}{3r}\, \sqrt{r^{2} - x^{2}}\, (r - x)\, dx = \frac{2h}{3} \int_{-r}^{r} \sqrt{r^{2} - x^{2}}\, dx - \frac{2h}{3r} \int_{-r}^{r} x \sqrt{r^{2} - x^{2}}\, dx. $$ The first integral is the area of a half-disk of radius $r$, namely $\frac{1}{2}\pi r^{2}$, while the second vanishes as the integral of an odd function over a symmetric interval. That is, $\text{Vol} = \frac{1}{3} \pi r^{2}h$.
Best Answer
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.