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let say we have a cone within a cylinder enter image description here

The volumes of cylinder and cone are related by $$ \frac{V_{cone}}{V_{cylinder}} = \frac{1}{3} $$

We have also following formulas for surface areas: $$ A_{cylinder} = 2\pi r(r+h) \\ A_{cone} = \pi r(r+ \sqrt{r^2 + h^2}) $$

with $r$ and $h$ being the radius and height.

Quastions:

  1. I am actually intrested if there is any simple relationship betwenn these sourfaces, like this: $$ A_{cone} = \lambda A_{cylinder} $$
  2. Also is following correct: If there are some $(r_0, h_0)$ which minimize the area of the cylinder, they will respectively minimize the area of the cone?

thanks

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  1. You already know the formulas for the two areas. By dividing them you see that the ratio is $$\frac{A_{\text{cone}}}{A_{\text{cylinder}}} = \frac{r+\sqrt{r^2+h^2}}{2(r+h)}$$There is no slick reduction that will somehow make that a constant. Just plugging in a few values of $r$ and $h$ will show you that it varies with respect to each.

  2. Technically this is correct, but only because the value of $r$ that minimizes $A_{\text{cylinder}}$ is $r = 0$ (and any $h$), which makes the area $0$, and of course it also makes $A_{\text{cone}} = 0$. I am sure that is not what you are after.

Perhaps you meant to minimize the area with respect to some condition?

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  • $\begingroup$ To 1) i did it actually already but i could not come up with a constant factor. so i thought there will be maybe some mathematical "trick" - e.g. some kind of transformation, wich could relate both surfaces as a factor. To 2) I am actually intrested in a proof of this concept. And yes the conditions given are: {r>= 0, h>=0, V_cone=v_0}. thanks $\endgroup$ – arash javan Jun 12 '19 at 9:27
  • $\begingroup$ Proof: The areas are $\ge 0$ so $0$ is the minimum value. $r =0$ makes both areas $0$ QED. $\endgroup$ – Paul Sinclair Jun 13 '19 at 3:03

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