# Finding endpoints for radius of cylinder when optimizing surface area

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 $$r \to \infty$$).

• 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. Nov 17, 2019 at 0:34
• @KM101 Would you have 3000/(pi * r) for that second term in surface area? What you're saying makes sense though Nov 19, 2019 at 3:11
• 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. Nov 19, 2019 at 4:09

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.