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You are asked to design a drink bottle out of a composite shape consisting of a right circular cylinder and a cone to contain a volume of 535 ml. In order to keep the cost of the product down, you must find the minimum surface area and dimensions for this bottle and prove this using calculus.

If need be the constraint that the cone's height is half of the cylinder can be added to the problem.

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  • $\begingroup$ You should add your own work on the problem. Otherwise the question will probably just be closed. $\endgroup$ – Henrik supports the community Sep 16 '17 at 13:34
  • $\begingroup$ is the volume of $$525ml$$ of both, cylinder and cone? $\endgroup$ – Dr. Sonnhard Graubner Sep 16 '17 at 13:38
  • $\begingroup$ Sorry, I should have clarified that the volume of both shapes had to be equal and as for the working out I only just started calculus and didn't know what to do. $\endgroup$ – Rupert Ebeling Sep 17 '17 at 10:05
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Well, assuming a cylinder with a cone on it's top.

  • The volume of a cone is given by: $$\mathscr{V}_\text{cone}=\frac{\pi\cdot\text{r}^2\cdot\text{h}}{3}\tag1$$
  • The volume of a cylinder is given by: $$\mathscr{V}_\text{cylinder}=\pi\cdot\text{r}^2\cdot\text{h}\tag2$$

So, for the volume of the bottle we get:

$$\mathscr{V}_\text{bottle}=\mathscr{V}_\text{cone}+\mathscr{V}_\text{cylinder}=\frac{\pi\cdot\text{r}^2\cdot\text{h}}{3}+\pi\cdot\text{r}^2\cdot\text{h}=\frac{4\cdot\pi\cdot\text{r}^2\cdot\text{h}}{3}\tag3$$

  • The surface area of a cone is given by: $$\mathscr{S}_\text{cone}=\pi\cdot\text{r}\cdot\sqrt{\text{r}^2+\text{h}^2}\tag4$$
  • The surface area of a cylinder is given by: $$\mathscr{S}_\text{cylinder}=2\cdot\pi\cdot\text{r}\cdot\left(\text{r}+\text{h}\right)\tag5$$

So, for the volume of the bottle we get:

$$\mathscr{S}_\text{bottle}=\mathscr{S}_\text{cone}+\mathscr{S}_\text{cylinder}=\pi\cdot\text{r}\cdot\sqrt{\text{r}^2+\text{h}^2}+2\cdot\pi\cdot\text{r}\cdot\left(\text{r}+\text{h}\right)=$$ $$\pi\cdot\text{r}\cdot\left\{\sqrt{\text{r}^2+\text{h}^2}+2\cdot\left(\text{r}+\text{h}\right)\right\}\tag6$$

Now, when we know $\mathscr{V}_\text{bottle}$, we can write:

$$\mathscr{V}_\text{bottle}=\frac{4\cdot\pi\cdot\text{r}^2\cdot\text{h}}{3}=\text{n}\space\Longleftrightarrow\space\text{h}=\frac{3\cdot\mathscr{V}_\text{bottle}}{4\cdot\pi\cdot\text{r}^2}\tag7$$

So, we can also write:

$$\mathscr{S}_\text{bottle}=\pi\cdot\text{r}\cdot\left\{\sqrt{\text{r}^2+\left(\frac{3\cdot\mathscr{V}_\text{bottle}}{4\cdot\pi\cdot\text{r}^2}\right)^2}+2\cdot\left(\text{r}+\frac{3\cdot\mathscr{V}_\text{bottle}}{4\cdot\pi\cdot\text{r}^2}\right)\right\}\tag8$$

Now, find:

$$\frac{\partial\mathscr{S}_\text{bottle}}{\partial\text{r}}=0\space\Longleftrightarrow\space\text{r}=\dots\tag9$$


In your example you will find:

$$\color{red}{\text{r}\approx0.0418521\space\text{m}\space\wedge\space\text{h}\approx0.0729175\space\text{m}}\tag{10}$$

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  • $\begingroup$ Thanks so much! Jan Eerland this was the answer I was looking for. $\endgroup$ – Rupert Ebeling Sep 17 '17 at 10:06
  • $\begingroup$ @RupertEbeling But my solution is not 100% right! So, I'm working on a good version $\endgroup$ – Jan Sep 17 '17 at 10:37
  • $\begingroup$ By not 100% right are you referring to the heights being the same my friend asked me if it were possible to do it with the heights being different as you would be able to make a smaller tsa with the measurements R = 4.60399999999971 Cone Height = 4.1179499998441 Cylender Height = 6.66136686089509 TSA ≈ 348.632 *Edit these measurements were obtained by brute force trying millions of different arrangements $\endgroup$ – Rupert Ebeling Sep 17 '17 at 11:30
  • $\begingroup$ @RupertEbeling I've made a document for you. Do you want to have it? $\endgroup$ – Jan Sep 17 '17 at 11:33
  • $\begingroup$ Yes please, Jan Eerland. $\endgroup$ – Rupert Ebeling Sep 17 '17 at 11:35

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