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Find radius of a cylinder with the biggest area surface inscribed in cone. Cone has radius R, and height H. (sorry for bad english) enter image description here

i've tried and got this: $x = (hr)/2(h-r)$, but what if h = r?

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  • $\begingroup$ What have you tried? $\endgroup$
    – Nosrati
    Nov 10 '18 at 16:20
  • $\begingroup$ Hello @3a43mka, welcome to MSE. Can you please add your own attempts in the question. $\endgroup$
    – Ernie060
    Nov 10 '18 at 16:20
  • $\begingroup$ Added my attempts> $\endgroup$
    – 3a43mka
    Nov 10 '18 at 16:42
  • $\begingroup$ I am assuming you are talking about only the curved surface area and not including the top and bottom of the cylinder? $\endgroup$ Nov 10 '18 at 17:13
  • $\begingroup$ Nope, full surface area $\endgroup$
    – 3a43mka
    Nov 10 '18 at 21:17
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So, I want you to imagine a cylinder inside a cone of radius $r$ and height $h$ and a cylinder having a radius say $x$. Now imagine that the cylinder has a certain height and I am going to ignore that, instead, I am going to take the remaining height and call it $h^{'}$ so we get the height of the cylinder as $h-h^{'}$. Now leave that aside for a minute. Now we will use the same relation you derived, $$\frac{h^{'}}{x} = \frac{h}{r}$$ So, $$h^{'} = \frac{hx}{r}$$ Now, the surface area $S_a$ is, $$ S_ a = 2 \pi x \Bigl(h-h^{'}\Bigl) + 2\pi x^2$$ which is equivalent to, $$ S_ a = 2 \pi x \Bigl(h-\frac{hx}{r}\Bigl) + 2\pi x^2 $$ which gives us, $$S_a = 2 \pi h \Bigl(x - \frac{x^2}{r}\Bigl) + 2\pi x^2 $$ Now in order to find the maximum area that can be described in the cone, we have to differentiate w.r.t $x$ and then set to $0$.
Now, $$\frac{dS_{a}}{dx}=0$$ for maximum area. $$2 \pi h \cdot \frac{d\Bigl(x - \frac{x^2 }{r}\Bigl)}{dx} + 2\pi \frac{dx^2}{dx}=0$$ Which gives us, $$2 \pi h \Bigl(1 - \frac{2x}{r}\Bigl) + 4\pi x=0$$ So, $$h\Bigl(1- \frac{2x}{r} \Bigl) + 2x = 0$$ and now using this relation you can find $x$ in terms of $h$ and $r$. Personally I would only use, the curved surface area becuase the solution is much nicer in that case, but whatever, to each his own.

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