Minimizing the volume of a cylinder using fixed surface area

Considering that I want my cylinder to have a fixed surface area of 0.3m^2, how can I minimize the volume of the cylinder. I have already tried to optimize it but I am only able to maximize the volume.

• Without additional information, this is a classic Lagrange multipliers problem. Jan 22, 2021 at 23:24
• How can I find the minimum volume using Lagrange multipliers? Jan 22, 2021 at 23:26
• Think of a cylinder with very small height. The infimum of the possible volumes of the cylinder is $0m^3$ Jan 22, 2021 at 23:29
• Can't you just think about the problem and realize the minimal volume is $0$? Try it! Jan 23, 2021 at 0:17

In the limits, the volume approaches 0, there technically isn't a minimum in the sense of 'local extrema'. Consider the following:

$$A=2\pi( r^2 + rh)=0.3$$

$$h =\frac{0.3}{2\pi r}-r$$

$$V=\pi r^2 h = \pi r^2 (\frac{0.3}{2\pi r}-r)=\frac{0.3r}{2} - \pi r^3$$

Notice as $$r \rightarrow 0$$, $$V \rightarrow 0$$, and of course $$h \rightarrow \infty$$.

You want to solve the problem:

$$\tag{P} \begin{cases} \text{minimize:} & V = \pi r^2 h \\ \text{u. c.:} & 2\pi (r^2 + r h) = 0.3 \\ & r,h > 0 \end{cases}$$

From the equality constraint you get:

$$\pi rh = 0.15 - \pi r^2$$

therefore your two-variables problem rewrites as one-variable optimization problem, i.e.:

$$\tag{P'} \begin{cases} \text{minimize:} & V = r (0.15 - \pi r^2) \\ \text{u. c.:} & 0 < r < \sqrt{\frac{0.15}{\pi}} \end{cases}\; .$$

Function $$V = r (0.15 - \pi r^2)$$ is of class $$C^\infty$$ and differentiating you find:

$$\begin{split} V^\prime &= 0.15 - 3\pi r^2 \\ V^{\prime \prime} &= -6\pi r \end{split}$$

hence $$V$$ is concave in $$]0,\sqrt{\frac{0.15}{\pi}}[$$ and takes its infimum in one of the extrema, i.e. either for $$r \nearrow \sqrt{\frac{0.15}{\pi}}$$ or for $$r\searrow 0$$; in either case you find $$V \to 0$$, which is not the mimimum because $$0$$ is not a value taken by function $$V$$ on $$]0,\sqrt{\frac{0.15}{\pi}}[$$.

Thus the minimum problem (P) has no solution at all.