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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.

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

2 Answers 2

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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 $h \rightarrow \infty$.

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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.

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