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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.

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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.


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. $$\max{\text{M}_{\space\text{T}}}_{\space7}\left(\frac{5}{2},\frac{3}{4}\right)<\frac{273\pi}{8}\approx107.206\tag{10}$$
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  • $\begingroup$ @ Jan Eerland thankyou for your solution! I am wondering how you obtained the r and h values in step/equation (5) and (6)? $\endgroup$ – this guy123 Oct 29 '17 at 0:25
  • $\begingroup$ @ Jan Eerland sorry about my above comment. I was thinking straight! Of course, those are the values from my question :) $\endgroup$ – this guy123 Oct 29 '17 at 0:40
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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 $7\frac{kg}{m^2} * SA \leq 7,500kg$

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  • $\begingroup$ thanks, you've made me understand the problem better. I will give it a go and come back here if i really cannot crack it. $\endgroup$ – this guy123 Oct 26 '17 at 3:27
  • $\begingroup$ Also, is it possible to then go and solve for the tank size (r and h)? $\endgroup$ – this guy123 Oct 26 '17 at 3:32
  • $\begingroup$ @thisguy123 the issue is it’s multivariable optimization so you are optimizing two variables with constraints. Is this a multivariable calculus course? $\endgroup$ – rb612 Oct 26 '17 at 3:36
  • $\begingroup$ No. I was originally thinking a solution might come from plotting a graph in which two curves would intersect, thus, giving an optimum point. But I see the issue here. The original question wants the maximum tank size for minimum tank cost.This may be as far as i can get. $\endgroup$ – this guy123 Oct 26 '17 at 3:40
  • $\begingroup$ @thisguy123 unless there is some way to eliminate one of the variables $r$ or $h$ by writing one in terms of the other, this is not single-variable optimization as far as I can tell. You're trying to optimize surface area, which is a function of $r$ and $h$. I'd be curious to know the solution though. $\endgroup$ – rb612 Oct 26 '17 at 6:15

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