ratio of the amount of tin...

**cutiemike1** · August 15th 2010, 02:52 PM

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.)

The Correct Answer: 0.95245

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..

I am glad to be corrected..Thanks for the help in advance..

**Vlasev** · August 15th 2010, 03:13 PM

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

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

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

**cutiemike1** · August 15th 2010, 03:35 PM

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!!

**Vlasev** · August 15th 2010, 03:42 PM

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

**cutiemike1** · August 16th 2010, 02:07 AM

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?

**Vlasev** · August 16th 2010, 03:34 AM

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.

**cutiemike1** · August 16th 2010, 01:05 PM

well...thanks again sir.. ^^

Thread: ratio of the amount of tin...

LinkBack

Thread Tools

Display

ratio of the amount of tin...

Similar Math Help Forum Discussions

Ratio for curve and point-ratio form

amount

rate & amount

mean amount?

Amount at the end of 4 years?

Search Tags