Minimizing the volume of a cylinder using fixed surface area

Question

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.

Answer 1

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 $h \rightarrow \infty$.

Answer 2

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.