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 $r \to \infty$).
Best Answer
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.