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I'm a student in Calculus class, and my teacher assigned us the following problem:

A cylinder is inscribed in a right circular cone of height $4$ and radius (at the base) equal to $6$. What are the dimensions of such a cylinder which has maximum volume?

The problem is I'm currently out of town, and the teacher is out of office. I've scoured my calculus book trying to find a similar problem to try and find a place to begin on this one, but I can't seem to find ANYTHING like it (a recurring issue in this class). I have a feeling it is an optimization problem but I honestly don't understand the question (how do you make a cylinder from a cone?).

Where do I even start with this problem? I feel that once I get a clearer understanding of what the question is asking, I'll be able to answer it on my own.

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  • $\begingroup$ @Moo, I see they have the answer there, but I still don't even understand the question. What is meant by "cylinder inscribed in a right circular cone"? How do you create a cylinder from a cone? $\endgroup$ – duper51 Nov 5 '16 at 4:41
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    $\begingroup$ "Inscribed in" means the cylinder just barely fits inside the cone. $\endgroup$ – David K Nov 5 '16 at 4:42
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    $\begingroup$ I think you are having trouble visualising this. Think of a can of beans standing upright. Then place a paper cone directly over the can so that the intersection is the circle at the top of the can. Sometimes it's better to start with the cylinder and put the cone round it! $\endgroup$ – Scott Burns Nov 5 '16 at 4:43
  • $\begingroup$ @DavidK, fantastic! This question makes much more sense now (definitely an optimization problem). Thanks so much for your help. $\endgroup$ – duper51 Nov 5 '16 at 4:44
  • $\begingroup$ @ScottBurns, that's a good way of putting it. I had a bit of trouble understanding what was meant by "inscribed", but now I see that it means "barely fit inside of", rather than "drawn/written on" $\endgroup$ – duper51 Nov 5 '16 at 4:46
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Below is a (not-to-scale and I-totally-didn't-misread-the-question-at-first-as-6-being-the-diameter-instead-of-the-radius) picture of the cone and cylinder in question (behold my abysmal drawing skills in Microsoft Paint thanks to my expired license on Mathematica!)

enter image description here

In any case, we can let the radius of the inscribed cylinder be $x$. That must therefore mean that the height of the cylinder is $4 - \frac{4}{6}x$. This makes the volume of the cylinder

$$V = \pi x^2 \left(4 - \frac{2}{3} x\right) = \pi \left(4x^2 - \frac{2}{3} x^3\right)$$

To find the maximum volume of the cylinder we take the derivative of $V$ with respect to $x$ and set it to $0$:

$$\frac{dV}{dx} = \pi \left(8x - 2x^2\right) = 0 \Rightarrow 8x - 2x^2 = 2x(4-x) = 0$$

$x = 0$ leads to a trivial case, which leaves us with $x = 4$. This means that the height is $4 - \frac{2}{3} \cdot 4 = \frac{4}{3}$, and the volume is therefore $$V = \pi \cdot 4^2 \cdot \frac{4}{3} = \frac{64}{3}\pi$$

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  • $\begingroup$ The radius is 6 i didnt get why you have written 3 $\endgroup$ – Archis Welankar Nov 5 '16 at 4:55
  • $\begingroup$ Oh! My apologies, I thought $6$ was the diameter. Let me correct that. (Do I really have to redraw that though...) $\endgroup$ – 2012ssohn Nov 5 '16 at 4:56
  • $\begingroup$ No if you just edit the maths I think I can understand it $\endgroup$ – Archis Welankar Nov 5 '16 at 4:58
  • $\begingroup$ I blame the 3-4-5 triangle for making me assume 6 was the diameter xD it's nicer when the hypotenuse is an integer! In any case, the corrections have been made, hope it helps! $\endgroup$ – 2012ssohn Nov 5 '16 at 5:01

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