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January 24

Napkin_ring_problem

I pretty much get it but I don't understand what is meant by "the smallest possible sphere" and why it (r) should matter. I thought that that was the whole point. I understand the cylinder vanishes when h = 2r but why for "the smallest possible sphere"? 196.213.35.146 (talk) 06:18, 24 January 2017 (UTC)[reply]

Well, if the sphere is narrower than the band, the band of the specified height cannot exist. For the "smallest possible sphere" the entire volume of the sphere is equal to the volume of the band, the entire diameter of the sphere is the width of the band, and the hole has no volume. Wnt (talk) 07:07, 24 January 2017 (UTC)[reply]

Yes I understand that perfectly but my primary question is not really answered. What is "the smallest possible sphere"? The only way I can make sense of it is that it is a sphere where r (and by implication h) is not equal to zero. The formula

V={\frac {\pi h^{3}}{6}}

would not work. Likewise for

V={\frac {4}{3}}\pi r^{3}.

196.213.35.146 (talk) 06:52, 25 January 2017 (UTC)[reply]

It means the smallest possible R given h. The problem does not specify the radius R, but it is specified that the depth of the hole is h; this is not possible unless R ≥ h/2. --catslash (talk) 10:56, 25 January 2017 (UTC)[reply]

Ah! Got it. Thank you. 196.213.35.146 (talk) 07:14, 27 January 2017 (UTC)[reply]

You can also look at this from the POV of volume. The band has volume

V={\frac {\pi h^{3}}{6}}

. When a sphere has diameter h, radius h/2, its volume is

V={\frac {4}{3}}\pi ({\frac {h}{2}})^{3}

... which works out to the same thing. If the sphere is smaller than that it has less volume than the band has to have. Maybe there is some amusing way to make the calculation work out if you suppose the hole has negative width? Wnt (talk) 01:06, 26 January 2017 (UTC)[reply]