# Thread: ratio of the amount of tin...

1. ## ratio of the amount of tin...

A cylindrical tin can holding 2gal. has its height equal to the diameter of its base. Another cylindrical tin can with the same capacity has its heights equal to twice the diameter of its base. Find the ratio of the amount of tin required for making the two cans with covers. (1gal = 231 cu. in.)

This is what I did, in order for me to get the ratio this is my working equation:

...and then the answer I got is r = 0.8818895. I don't which part am I wrong in solving the ratio..

2. Your mistake is that you are using only one copy of the base area. You need two since you are including the lids! Just use twice the base area and you'll get your answer. However, I think you are making it unnecessarily difficult for yourself by doing it the way you seem to have done it.

Here's a better way. Let the radius of the small can be $r_1$ and that of the big can $r_2$. Since they have the same capacity, I will measure the smallness by their height. That is the one where the height is equal to the diameter is the small one and the one with twice its diameter is the big one.

So the small one has diameter $2r_1$, so its height is $r_1$. Its volume is $V_1 = \pi r_1^22r_1 = 2 \pi r_1^3$.
The big one has diameter $2r_2$, so its height is $2(2r_2) = 4r_2$ and its volume is $V_2 = \pi r_2^24r_2 = 4\pi r_2^3$.

Since the volumes are the same you have

$2\pi r_1^3 = 4\pi r_2^3$ which means $r_1^3 = 2 r_2^3$ after cancellation. This means $r_1 = 2^{1/3}r_2$

Now for you equation. The ratio should be

$\displaystyle r = \frac{2BaseArea_1+SideArea_1}{2BaseArea_1+SideArea _1} = \frac{2\pi r_1^2+2\pi r_1 (2r_1)}{2\pi r_2^2 + 2\pi r_2 (4r_2)}$

$\displaystyle= \frac{\pi r_1^2(2+4)}{\pi r_2^2(2+8)} = \frac{6r_1^2}{10r_2^2} = \frac{3.2^{2/3}r_2^2}{5r_2^2} = \frac{3.2^{2/3}}{5} = 0.9524$

3. Originally Posted by Vlasev
Your mistake is that you are using only one copy of the base area. You need two since you are including the lids! Just use twice the base area and you'll get your answer. However, I think you are making it unnecessarily difficult for yourself by doing it the way you seem to have done it.
w0w!.haha.so in short it is just like finding the TSA..thanks you so much... i mislooked that one..thank you thank you!!

4. No problem. It just seems that you have overlooked the small detail since you were using some more complicated expressions.

5. Originally Posted by Vlasev
No problem. It just seems that you have overlooked the small detail since you were using some more complicated expressions.
Yes sir! But do you think, what I just did is very complicated?

6. Well, it's not VERY complicated, but it's overly complicated, considering that when I wrote it out it took a lot less symbols and fractions. I mean, imagine punching it all in a calculator! Although, it does seem like you have done the correct calculation, since considering your formula, that's the answer one gets, so I retract my statement that the mistake you made is because of the complicated expression. Your mistake was in the defining expression (you missed the lid) and not afterwards.

7. well...thanks again sir.. ^^