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.
Best Answer
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!)
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 $$V = \pi \cdot 4^2 \cdot \frac{4}{3} = \frac{64}{3}\pi$$