3
$\begingroup$

Given a cube of side $a$, what is the volume of the greatest cylinder that can be inscribed in it, such that its axis coincides with the longest diagonal of the cube?

It is quite obvious that for obtaining maximum volume, we would have to stretch the cylinder until it touches the cube’s faces at six distint points, but is that even possible? I’ll explain what I mean by converting this (sort of) to a $2D$ problem.

Let’s say I have a rectangle $ABCD$, and inside the rectange is another (smaller) rectangle with its axis being the diagonal $AC$. (axis here refers to the line passing through the center and parallel to any one pair of sides). Now, intuitively or even by drawing a picture it’s evident that it shouldn’t be possible to have all four corners of the smaller rectangle touching the bigger rectangle, leaving one corner hanging in the air. Is there a way to prove/disprove this? It’s quite possible in the case of a square, though.

What I did here was take a side-view of the original configuration, where the bigger rectangle is determined by the farthest edges (of length $a$) of the cube and the shorter diagonals (length $\sqrt 2 a$).

I’m looking for is way to relate the height and radius of the cylinder, after which the task is trivial using calculus.

$\endgroup$
6
  • $\begingroup$ If the inscribed cylinder’s axis is a long diagonal, then it will touch at six points, not four: three faces converge symmetrically at each vertex. That symmetry is key: if you the faces that converge at either end of the diagonal, you get an equilateral triangle. $\endgroup$
    – amd
    Commented Apr 17, 2020 at 22:21
  • $\begingroup$ @amd Yes, my bad. Can you explain your last statement? $\endgroup$
    – Vishu
    Commented Apr 17, 2020 at 22:29
  • $\begingroup$ Left out a bunch of words. Sorry. It should read “if you intersect the faces that converge at either end of the diagonal with a plane perpendicular to the diagonal, you get an equilateral triangle.” $\endgroup$
    – amd
    Commented Apr 17, 2020 at 22:34
  • $\begingroup$ @amd and how would that help? $\endgroup$
    – Vishu
    Commented Apr 17, 2020 at 22:48
  • $\begingroup$ The circle inscribed in that triangle is an end cap of the cylinder. It also shows that each end of the cylinder contacts the cube at three points. $\endgroup$
    – amd
    Commented Apr 17, 2020 at 22:49

2 Answers 2

Reset to default
3
+50
$\begingroup$

Position the cube so that a vertex is at the origin and the cube lies in the first octant. The long diagonal has length $a\sqrt3$, so if the cylinder’s height is $h$, its near cap lies at a distance of $\frac12\left(a\sqrt3-h\right)$ from the origin. The normal to this cap’s plane is $(1,1,1)$, so an equation of this plane is $$x+y+z=\frac{\sqrt3}2\left(a\sqrt3-h\right).$$ The expression on the right-hand side is the axis-intercept of this plane with all three coordinate axes. The end cap touches the $x$-$y$ plane at the midpoint of the $x$- and $y$-intercepts, so using the Pythagorean theorem, we can obtain the square of its radius, namely $$\frac38\left(a\sqrt3-h\right)^2-\frac14(a\sqrt3-h)^2 = \frac18\left(a\sqrt3-h\right)^2.$$ The rest of the task, as you say, is a trivial calculus exercise.

$\endgroup$
3
$\begingroup$

Following amd’s comment consider a plane containing the cup circle (of radius $r$) of the cylinder. The plane cuts a pyramid (of height $h$) from the cube. Let $V$ be the volume of the pyramid and $S$ be the area of its base, which is an equilateral triangle with a side $s$. Then $V=\frac 13 Sh=\frac {\sqrt 3}{12}s^2h$. On the other hand, $V=\frac {d^3}6$, where $d$ is the length of the part of an edge of the cube cut by the plane. Also we have $s^2=2d^2$. It follows $V=\frac {\sqrt 3}{12}s^2h=\frac {s^3}{12\sqrt{2}}$ and so $s=h\sqrt{6}$. The circle inscribed into the triangle has radius $r=\frac s{2\sqrt{3}}=\frac h{\sqrt 2}$. The cyliner has height $H=a\sqrt{3}-2h$ and the volume $$\pi r^2H=\pi\frac {h^2}{2}\left(a\sqrt{3}-2h \right).$$

$\endgroup$
4
  • $\begingroup$ How is $V=\frac{d^3}{6}$? Is there some formula I’m not seeing? $\endgroup$
    – Vishu
    Commented Apr 20, 2020 at 20:37
  • $\begingroup$ We have $V=\frac {d^3}6$ because the edges of the pyramid incident to the vertex $A$ of the cube are mutually perpendicular and have length $d$. So an area of any face incident to $A$ equals $\frac {d^2}2$ and the height of the pyramid perpendicular to the face has length $d$. $\endgroup$
    – Alex Ravsky
    Commented Apr 21, 2020 at 1:34
  • $\begingroup$ Nice. Can any face of a pyramid be taken as its ‘base’ for volume calculations? $\endgroup$
    – Vishu
    Commented Apr 21, 2020 at 7:32
  • $\begingroup$ @Tavish Yes, if all the faces of the pyramid are triangles. $\endgroup$
    – Alex Ravsky
    Commented Apr 21, 2020 at 7:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .