Minimizing costs of a specific geometry shape

Question

I have geometry mathematical problem. I have a shape that is made of a cylinder and two half spheres as to top and bottom. How can I minimize the cost of this shape when the volume is known and the cylinder part costs $k_1$ dollar per square meters and the sphere part costs $k_2$ dollar per square meter.

Answer 1

Well, the volume of that shape is given by:

$$\mathcal{V}=\mathcal{V}_\text{cylinder}+\mathcal{V}_\text{hemisphere}+\mathcal{V}_\text{hemisphere}=\mathcal{V}_\text{cylinder}+\mathcal{V}_\text{sphere}\tag1$$

Now, for the volume of a cylinder:

$$\mathcal{V}_\text{cylinder}=\pi\cdot\text{r}_\text{cylinder}^2\cdot\text{h}\space_\text{cylinder}\tag2$$

And for the sphere:

$$\mathcal{V}_\text{sphere}=\frac{4}{3}\cdot\pi\cdot\text{r}_\text{sphere}^3\tag3$$

Now, we also know that:

$$\text{r}=\text{r}_\text{cylinder}=\text{r}_\text{sphere}\tag4$$

So the total volume is given by:

$$\mathcal{V}=\pi\cdot\text{r}^2\cdot\text{h}\space_\text{cylinder}+\frac{4}{3}\cdot\pi\cdot\text{r}^3=\frac{\pi\text{r}^2}{3}\cdot\left(3\text{h}\space_\text{cylinder}+4\text{r}\right)\tag5$$

And the surface area is given by:

$$\mathcal{S}=2\cdot\pi\cdot\text{r}\cdot\text{h}\space_\text{cylinder}+4\cdot\pi\cdot\text{r}^2\tag6$$

And for the costs we can set up a function:

$$\mathcal{K}=\text{K}_1\cdot2\cdot\pi\cdot\text{r}\cdot\text{h}\space_\text{cylinder}+\text{K}_2\cdot4\cdot\pi\cdot\text{r}^2\tag7$$

Well, you know the volume:

$$\mathcal{V}=\frac{\pi\text{r}^2}{3}\cdot\left(3\text{h}\space_\text{cylinder}+4\text{r}\right)\space\Longleftrightarrow\space\text{h}\space_\text{cylinder}=\frac{\mathcal{V}}{\pi\text{r}^2}-\frac{4\text{r}}{3}\tag8$$

So, in order to minimize the costs we can write:

$$\frac{\partial\mathcal{K}}{\partial\text{r}}=\frac{\partial}{\partial\text{r}}\left\{\text{K}_1\cdot2\cdot\pi\cdot\text{r}\cdot\left(\frac{\mathcal{V}}{\pi\text{r}^2}-\frac{4\text{r}}{3}\right)+\text{K}_2\cdot4\cdot\pi\cdot\text{r}^2\right\}=$$ $$8\text{K}_2\pi\text{r}-\frac{16\text{K}_1\pi\text{r}}{3}-\frac{2\text{K}_1\mathcal{V}}{\text{r}^2}=0\space\Longrightarrow\space\text{r}=\left(\frac{1}{4}\cdot\frac{\text{K}_1\mathcal{V}}{\pi\text{K}_2-\frac{2\pi\text{K}_1}{3}}\right)^\frac{1}{3}\tag9$$

And so the height will be:

$$\text{h}\space_\text{cylinder}=2\cdot\left(\frac{6}{\pi}\right)^\frac{1}{3}\left(\frac{\text{K}_2}{\text{K}_1}-1\right)\cdot\left(\frac{\text{K}_1\mathcal{V}}{3\text{K}_2-2\text{K}_1}\right)^\frac{1}{3}\tag{10}$$