I asked a question about this problem previously but for this post I am asking a different question about the same problem.


The volume of a cylinder equals 𝑉 cubic inches, where 𝑉 is a constant. Find the proportions of the cylinder that minimize the total surface area.

I know how to get the answer to this problem. What I have trouble with is visualizing what the graph of $\frac{dS}{dr}$ is if $S(r)$ is the total surface area as a function of the radius. The equation for $\frac{dS}{dr}$ is $\frac{dS}{dr}=\frac{4\pi r^3-2V}{r^2}$ and since $V=\pi r^2h$, shouldn't the total surface area be a function of both the radius and the height, so basically $S(r,h)$.

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    $\begingroup$ You can use a $3$ dimensional surface with axes $S,r,h$. Why not ask the same question for $V$ itself which depends on both $r$ and $h$? $\endgroup$ – Peter Foreman Jun 30 '19 at 16:16
  • $\begingroup$ What you want to graph $S(r,h)$ or dS/dr ? $\endgroup$ – iaeliyen Jun 30 '19 at 16:29
  • $\begingroup$ @AjayMishra $\frac{dS}{dr}$, but how do I graph this if I don't know what $V$ is? $\endgroup$ – user532874 Jun 30 '19 at 16:39
  • $\begingroup$ @user532874 see my answer. $\endgroup$ – iaeliyen Jun 30 '19 at 16:45

Green axis : h, Red axis: r Here, Green axis : $h$, Red axis: $r$

Since, $$\cfrac{dS}{dr} = \cfrac{4 \pi r^3 - 2V}{r^2} = \cfrac{4 \pi r^3 - 2 \pi r^2 h}{r^2} = 2 \pi ( 2r - h) $$ If I define $z = \cfrac{dS}{dr}$ ,then $$ z = 2 \pi (2r-h) $$ is just equation of a plane.

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  • $\begingroup$ How can $\frac{dS}{dr}$ have just $dr$ on the bottom when both $r$ and $h$ are inputs? $\endgroup$ – user532874 Jun 30 '19 at 16:49
  • $\begingroup$ Oh, it is just $ \cfrac{ \partial S}{ \partial r}$ . Are you familiar with partial derivatives? $\endgroup$ – iaeliyen Jun 30 '19 at 16:53
  • $\begingroup$ No I will have to learn that thanks $\endgroup$ – user532874 Jun 30 '19 at 17:06
  • $\begingroup$ That is derivative with respect to r and treating h as constant. There is nothing wrong, here. That often occurs in multivariable calculus $\endgroup$ – iaeliyen Jun 30 '19 at 17:09
  • $\begingroup$ That's a partial derivative? $\endgroup$ – user532874 Jun 30 '19 at 17:12

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