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.


  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?


  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?

  • $\begingroup$ 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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