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I was working on an optimization problem for minimizing the surface area of a cylinder given that the volume is fixed at $1500 cm^3$. I found the critical point and classified it as a minimum of the surface area, but I also want to check the endpoints for $r$, the radius of the cylinder.

How do I find the smallest and largest values of $r$ (the endpoints over $r$ where we can have global extrema)? Intuition tells me that the smallest value of $r$ is $0$, and the largest value of $r$ is $\infty$, but if I have these numbers for the radius, I would not be able to constrain the volume to $1500 cm^3$ (i.e. I wouldn't be able to find the height of the cylinder that corresponds to $r = 0$ or $r \to \infty$).

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  • $\begingroup$ Clearly, you can't have $R = 0$, since it makes no sense physically. Also, the equation for surface area with respect to radius becomes $A_{\text{surface}} = 2\pi R^2+\frac{3000}{R}$, which is decreasing on $R \in \left(0, \sqrt[3]{\frac{750}{\pi}} \right]$ and increasing on $R \in \left[\sqrt[3]{\frac{750}{\pi}}, +\infty \right)$. So, while there isn't any limit to how large or how small ($R \to 0^+$) you want $R$ to be, you're just increasing the surface area, which is exactly the opposite of what you want to do here. $\endgroup$
    – KM101
    Nov 17, 2019 at 0:34
  • $\begingroup$ @KM101 Would you have 3000/(pi * r) for that second term in surface area? What you're saying makes sense though $\endgroup$
    – Buddhapus
    Nov 19, 2019 at 3:11
  • $\begingroup$ Nope, since the second term corresponds to $2\pi Rh$, and $V = \pi R^2h \iff R = \frac{V}{\pi R^2}$. Plugging this in the second term simplifies to give $\frac{2V}{R}$. The $\pi$ factors cancel out. $\endgroup$
    – KM101
    Nov 19, 2019 at 4:09

1 Answer 1

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The formula for the volume of a cylinder is:

$$V = \pi r^2h$$

As was stated in the comments, it doesn't make sense to have a radius of $0$, so that option is eliminated. Let's start by ignoring the surface area for now and just focusing on the volume. We can start by plugging in our volume.

$$1500 = \pi r^2 h$$ $$\frac{1500}{\pi h}=r^2$$ $$\sqrt{\frac{1500}{\pi h}} = r$$

We don't need to include the $\pm$ since the radius is always positive.

We use the chain rule to find the derivative:

$$-\dfrac{5\sqrt{15}}{\sqrt{{\pi}}h^\frac{3}{2}} = r'$$

You will quickly notice, however, that this function has no zeros. Thus, $r$ has no critical points and no global extrema.

Therefore, $r$ can approach both $0$ and $\infty$ without changing the cylinder's volume (as long as the height changes with it according to the above equation for $r$).

Surface area is a different story, though. As was stated by KM$101$ in the comments, you would just increase the surface area as $r \to \infty$, which is not what the problem asked for.

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