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I have been sitting here staring at this problem for hours and I have no idea where to start..... the problem reads:

Imagine a flat bottomed cylindrical pot with a radius of $4$. A marble with a radius $0 < r < 4$ is placed in the bottom of the pot. What is the radius of the marble that requires the most water to cover it completely?

I really hope somebody can help me, please!

Vedran Šego
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cschurman
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  • Is there a hint with the problem (I found it in the book)? – Amzoti Jul 22 '13 at 02:32
  • No, @Amzoti: the phrase 0<r<4 blocked everything after 0. Using "run on" <x< seems to do that. x can be anything. – amWhy Jul 22 '13 at 02:34
  • Perhaps I'm naive, but wouldn't it just be the largest possible surface area that can still fit in the pot? – rurouniwallace Jul 22 '13 at 02:37
  • the radius of the marble has to be greater than 0 but can't be bigger than 4. that is what i meant by 0<r<4 – cschurman Jul 22 '13 at 02:37
  • it is an "optimization" calculus problem. and no because the water has to cover the entire marble. if that marble was the largest possible surface area that can fit in the pot, the water wouldn't be able to cover the entire marble. – cschurman Jul 22 '13 at 02:39
  • @cschurman: The meaning of your statement about the marble's radius is clear. The problem is that, the way text-handling works on sites like this, some combinations involving the less-than sign can be interpreted by the system as formatting symbols. So, in the original version of your question, all the stuff after the "0" failed to appear. (This is a good reason to use the "TeX" formatting for the mathematical portions of a question: enclose equations within dollar signs.) – Blue Jul 22 '13 at 02:46
  • so how would i need to right it? I guess i'm just not following what you are meaning... – cschurman Jul 22 '13 at 02:48
  • @cschurman: It's okay now. amWhy fixed it. If you click on the "edited [x mins] ago" link, you can see the changes; notice the dollar signs. (Somewhere, there's a nice little introduction to "TeX" ---the way we write math here--- but I can't seem to find the link at the moment.) – Blue Jul 22 '13 at 02:54

2 Answers2

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I presume you know a little calculus? If so then find the maximum of Vol(water) = Vol(cylinder) - Vol(sphere), where the depth of the cylinder is the same as the diameter of the sphere.

MikeFee
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  • i do, i am taking this class, and it is over in a week, and i am falling way behind, and i'm lost on ALOT of things, wat do you mean by this, after looking at it some more, i was thinking i need to find the circumference of the marble, and then the volume of the cylinder? – cschurman Jul 22 '13 at 02:45
  • @cschurman $V(r)=16\pi{r}-\frac{4}{3}\pi{r}^3$. Find the $r$ that minimizes $V$. – rurouniwallace Jul 22 '13 at 02:47
  • to do so, I would set the equation equal to zero, then take 16 and multiply it by pi, then multiply 4/3 by pi, and be left with 50.27r-4.19r^3=0, then would i solve for r? – cschurman Jul 22 '13 at 02:53
  • @cschurman no there's much more to it than that. You need to take the derivative, find the relative extrema, and evaluate the function at the critical points to find which is the max. – rurouniwallace Jul 22 '13 at 02:56
  • @cschurman Also, I must make a correction. That should be $V(r)=16\pi{(2r)}-\frac{4}{3}\pi{r}^3$. Sorry for the confusion. – rurouniwallace Jul 22 '13 at 02:57
  • which is another part that i'm lost at. I am falling way behind in this class, and this assignment is due tomorrow, and I just can't seem to get any of it figured out – cschurman Jul 22 '13 at 02:57
  • @cschurman Read this article, it does a very good job at explaining optimization. If you still don't know what to do after reading that, please edit your question with details on what specifically it is that's tripping you up, and we'll help as best as we can. – rurouniwallace Jul 22 '13 at 03:09
  • @ZettaSuro what is the equation that I am trying to maximize? that article is kind of helping me with this! – cschurman Jul 22 '13 at 03:23
  • @cschurman This one: $V(r)=16\pi{(2r)}-\frac{4}{3}\pi{r}^3$. – rurouniwallace Jul 22 '13 at 03:25
  • @ZettaSuro what would be my constraint equation? – cschurman Jul 22 '13 at 03:27
  • @ZettaSuro what would V" be? – cschurman Jul 22 '13 at 03:44
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The volume of water needed to cover a marble of radius $r$ is te volume of the cylinder of water minus de volume of the marble, this is $V(r) = \pi(4)^2(2r)-\frac{4}{3}\pi r^3 = 32\pi r - \frac{4}{3}\pi r^3$.

We want the maximum of $V$, since $\displaystyle\frac{dV}{dr} = 32\pi - 4\pi r^2$ we have $\displaystyle\frac{dV}{dr} = 0$ if $32\pi - 4\pi r^2=0$, then $r=\sqrt{8}=2\sqrt{2}$. You should verify that $2\sqrt{2}$ is a maximum, this is, show that $V''(2\sqrt{2})<0$.

Cure
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  • so would V′′(22√)<0 be the verification, or do i have to go farther into verifying? – cschurman Jul 22 '13 at 03:32
  • Remember that if $V'(a)=0$ we only know that $a$ is a critical point, but we know that $a$ its a maximum if $V''(a)<0$ and a minimum if $V''(a)>0$, then it will be enough to show that $V''(2\sqrt{2})<0$. Also remember that $V''$ means $\frac{d^2V}{dr^2}$, this is the second derivative of $V$ or the first derivative of $V'$, derive again $V'$ and you have $V''$, then verify that $V''(2\sqrt{2})<0$ – Cure Jul 22 '13 at 04:04