# Optimization Metal Tank Problem

1. Jan 4, 2006

### Xcron

Ok, well the problem states:

A metal storage tank with volume V is to be constructed in the shape of a right circular cylinder surmounted by a hemisphere. What dimensions will require the least amount of metal?

The first step that I took was to draw a picture. I just drew a semicircle with a right circular cylinder right in the middle with the base on the bottom of the hemisphere and the top touching at two points to the rounding of the semicircle. I've attached a crude representation of my drawing.

Then, I defined the volume of this storage tank, V = (1/2)(4/3*pi*r(hemisphere)^3) - (pi*r(cylinder)^2*h(cylinder)). I kept in mind the fact that V is a constant that was given in the problem and that I would use it in my final equation that I would optimize.

I am not sure of this is correct but it seems to me that the metal storage tank would be a hemisphere with a hole that is a right circular cylinder...

Next, I tried to eliminate one of the three variables. I did this by drawing a triangle from the top-right corner of the cylinder to the center of the cylinder and connected it to the right side of the cylinder. I have attached a representation of this too.

I used the pythagorean theorem from there and said that (h/2)^2 = (r(hemi))^2 - (r(cyli))^2. Then I continued to square root both sides to solve for h/2.

I thought that I would need to solve for the surface area of the tank to see how much material would be needed and I write an equation for that as A = (1/2)(4*pi*r(hemi)^2) - (2*pi*r(cyli)^2 + 2*pi*r(cyli)*h). After that I tried to get r(cyli) onto one side so that I could solve for it but I ended up with a rather large and extremely convoluted mess from which I was not able to do so.

Any help would be appreciated. I'm stuck as to the actual representation of the problem...*sigh*...

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2. Jan 4, 2006

### Staff: Mentor

I think you have the geometry wrong. It sounds to me like the problem is describing an old grain-silo type shape. Or a long cylinder with a rounded end kind of like a condom if you haven't seen old grain silos. Sorry I can't think of a better description. I think that the radius of the hemisphere is equal to the radius of the cylinder and that the volumes and surface areas are simply added.

So you have a cylinder with a hemisphere stuck on top. I think there are only two design variables, the radius and the height of the cylinder portion. You should then be able to solve for e.g. r in terms of V and h and then substitute into the equation for the surface area to get A as a function of h and V.

-Dale

Last edited: Jan 4, 2006
3. Jan 4, 2006

### Mindscrape

Why did you subtract the cylinder's volume? And yes, I think it's fair to assume the radius is the same for the two, which now makes it an easy/straightforward optimization problem.

4. Jan 5, 2006

### Xcron

Hhahahahahahah! If you are correct then it will truly become an easy and straightforward problem. I must admit that I did not understand the visualization part of the problem correctly...perhaps because I have not been trained to deal with images of such words as surmounted.

Is that what surmounted really means by any chance?....Because I did a problem where the image was once given and it said a rectangle that is surmounted by a semicircle and it looked just like what I was trying to draw out except there was a rectangle instead of a cylinder.

If only the said had stated that it was mounted on top of a standing cylinder instead...*sigh*...

I subtracted the cylinder's volume because I thought that is where something would be stored...this idea was produced from my confusion after trying to visualize the problem...I'm sorry.

Would anyone confirm as to the explicit meaning of the imagery/language of the problem?

I believe that DaleSpam has quite a good set-up but I am not sure if that is the correct one that I should be using...

Also, would anyone explain how I may use the mathematical pictures of pi and such things that I have seen people use instead of my crude typing of i.e. 2*pi*r(hemi)...

Last edited: Jan 5, 2006
5. Jan 5, 2006

### Mindscrape

But you wouldn't subtract it for storage because then you would have a rod with hemisphere dug into it. I think they want the volume of the whole thing anyway, for possible storage. So you probably had it right, except add this time.

$$V = \frac{2}{3} \pi r^3 + \pi r^2 h$$

Oh, the typesetting is TeX. Ever used LaTeX before? Pretty much what this is. If you ever get a math project where you have to type something up, then you should use LaTeX.

Last edited: Jan 5, 2006
6. Jan 5, 2006

### Staff: Mentor

Am I allowed to confirm that I am correct.
Check out the LaTeX thread in the tutorials section (https://www.physicsforums.com/showthread.php?t=8997) It has a bunch of examples and "how to" links and you are allowed to practice there. If you practice there please be sure to delete your post.

-Dale