How to optimise the size of a cylinder/tank?

Question

Given/Known data:

Maximum mass of the tank must not exceed 7500 kg. Maximum size of the cylinder: diameter of 1.5m, height of 2.5m. Tank will carry water. Made of steel (plates): $Cost_0 = {$}75/m^2$; $Mass = 7 \frac {kg}{m^2}$

I would like to find the maximum size of the tank but for minimum cost using this data. I am going to ignore any other costs except for material cost. So my working so far: I know the total surface area of a cylinder is: $$SA = 2*pi*r^2 +2*pi*r*h$$

Therefore, the cost would be: $$ Cost = SA * Cost_o$$

This is where I am stuck. Is it possible to get one answer or is there multiple? I am unsure where to go next. Any help is appreciated. This is all the data I have.

Thanks.

Answer 1

Assuming a complete closed cylinder, the volume and surface area are given by:

• $$\mathscr{V}\left(\text{h},\text{r}\right)=\pi\cdot\text{r}^2\cdot\text{h}\tag1$$
• $$\mathscr{S}\left(\text{h},\text{r}\right)=\pi\cdot\text{r}^2+\pi\cdot\text{r}^2+2\pi\cdot\text{r}\cdot\text{h}=2\pi\cdot\text{r}\cdot\left(\text{h}+\text{r}\right)\tag2$$

For the costs we can write:

$$\text{C}_{\space\text{n}_1}\left(\text{h},\text{r}\right)=\text{n}_1\cdot\mathscr{S}\left(\text{h},\text{r}\right)=2\pi\cdot\text{r}\cdot\text{n}_1\cdot\left(\text{h}+\text{r}\right)\tag3$$

Where $\text{n}_1$ is the cost per square meter.

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Now, let's set the things we know:

1. Let $\text{n}_2=7$ be the weight per square meter material, then the total mass is given by: $${\text{M}_{\space\text{T}}}_{\space7}\left(\text{h},\text{r}\right)=\text{n}_2\cdot\mathscr{S}\left(\text{h},\text{r}\right)=2\pi\cdot\text{r}\cdot7\cdot\left(\text{h}+\text{r}\right)<7500\space\Longleftrightarrow\space$$ $$\text{r}\cdot\left(\text{h}+\text{r}\right)<\frac{3750}{7\pi}\approx170.523\tag4$$
2. $$\text{r}<\frac{3}{4}\tag5$$
3. $$\text{h}<\frac{5}{2}\tag6$$

Now, we can find out what the maximum values are for the volume, surface area, costs and mass:

1. $$\max\mathscr{V}\left(\frac{5}{2},\frac{3}{4}\right)<\frac{45\pi}{32}\approx4.418\tag7$$
2. $$\max\mathscr{S}\left(\frac{5}{2},\frac{3}{4}\right)<\frac{39\pi}{8}\approx15.315\tag8$$
3. $$\max\text{C}_{\space75}\left(\frac{5}{2},\frac{3}{4}\right)<\frac{2925\pi}{8}\approx1148.644\tag9$$
4. $$\max{\text{M}_{\space\text{T}}}_{\space7}\left(\frac{5}{2},\frac{3}{4}\right)<\frac{273\pi}{8}\approx107.206\tag{10}$$

Answer 2

You have correctly identified both equations. Think about what you really need to be optimizing in the second equation. You have some constant cost per square meter. You are multiplying that by the surface area. So to minimize total cost, you want to minimize your surface area, right? So minimizing your surface area will minimize your cost automatically because cost is just a multiple of your total surface area.

So you want to minimize your cost while also fulfilling the constraints of the tank. You need the constraint that your total mass cannot exceed 7500kg, and every square meter of surface area adds 7kg to your mass. What is a bit ambiguous is if we are accounting for the water mass inside the tank (I'm presuming not because the mass is given in units meters squared, which is area, not volume).

In that case, you can say $7\frac{kg}{m^2} * SA \leq 7,500kg$