anonymous
  • anonymous
A cylindrical can of soda is 12cm tall and has a diameter of 6cm. The box for a 12 pack of soda has a length of 25cm, width of 19cm, and height of 12.2cm. The cans are placed in the box in three rows of 4. What is the volume of space that is not used when a 12-pack box of soda is full? Show two or more ways of solving this problem.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
bahrom7893
  • bahrom7893
|dw:1371334394203:dw|
bahrom7893
  • bahrom7893
|dw:1371334439569:dw|
bahrom7893
  • bahrom7893
What's the volume of the can? Sorry my cylinder was drawn horribly there.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
V=pi(r)^(h) V=pi(3)^(12) V=339.12cm cubed for 1 can so x12 cans = 4069.44 cm cubed....Right???
kirbykirby
  • kirbykirby
No the volume is the Area of the base * height... V = pi*r^2 * h
kirbykirby
  • kirbykirby
wait your number seems correct when i calculated it.. but I don't know how you got it from pi*r^h??
anonymous
  • anonymous
Volume of the box minus total volume of the cans.\[\left(25*19*\left(12+\frac{2}{10}\right)\right)-12((\pi 3{}^{\wedge}2)12)=5795-1296 \pi =1723.5\]
kirbykirby
  • kirbykirby
You can do this a 2nd way as the problem says, but it would be more tedious (at least the way I'm thinking of it). You could imagine the cans all touching each other in a rectangular array and imagine a circumscribed box over them. Then, you'd calculate the volume of that small box minus the total value of the cans, and the add up the residual volume left in the "larger box"
anonymous
  • anonymous
Normally in Mathematics any initial solution is good enough.
kirbykirby
  • kirbykirby
I agree. Unfortunately, the question doesn't agree >_>
anonymous
  • anonymous
Thank you for the medal.
kirbykirby
  • kirbykirby
np =]
anonymous
  • anonymous
Thanks for your help...that's what I got BUT I still need to find a second method to solve the problem :(
kirbykirby
  • kirbykirby
You can approach it the way I described it... Let me draw as best as I can
kirbykirby
  • kirbykirby
|dw:1371337202013:dw|
kirbykirby
  • kirbykirby
now imagine there are three rows of cans like that... then imagine a box that just covers the cans... as shown in dotted lines... You can then find the area of that dotted box minus the volume of all cans... then you calculate the volume of the remaining white space... where there;s a length of 1 cm and 0.2cm left... and the width at the back remaining would be 1 cm since there would be 3 cans of 6 cms = 18cm (out of 19cm width)
kirbykirby
  • kirbykirby
It's really just a longer way of doing the same thing in essence, but I would still count it as a valid 2nd method.
anonymous
  • anonymous
we can mix area calulations with volume? Do you have it in an equation? Thanks
kirbykirby
  • kirbykirby
Well a volume is really taking an area, and multiplying another dimension to it. Anyway, I only drew the face of the front of the box, you can draw a side view showing the 3 cans along the width
kirbykirby
  • kirbykirby
Well to find the volume remaining from the cans and the dotted box: The dotted box has volume 24*12*18 = 5184 The volume of all the cans = 1296*pi as found before The volume remaining in between is 1296*pi-5184 = 1112.5 Now the remaining volume between the large box and the small dotted box: large box has volume 25*12.2*19 = 5795 so the volume remaining is Large box - Small box = 5795 - 5184 = 611 So, we add up the 2 remaining volumes: 1112.5+611=1723.5
kirbykirby
  • kirbykirby
A third method: Instead of find the residual volume between the dotted box and large box by doing subtraction... you could separate the remaining space into 3 rectangular prism spaces and you can calculate each of those volumes separately
kirbykirby
  • kirbykirby
So you can calculate those 3 sections as: (hopefully you can visualize them based on the dimensions:) Section1 (block on the right): 1*19*12.2 = 231.8 Section2 (block on top): 24*19*0.2 = 91.2 Section 3 (black behind the cans): 1*12*24=288 231.8 + 91.2 + 288 = 611 as we found before
anonymous
  • anonymous
I understand how your concept but how did you get 1112.5 from 1296*pi-51840?
anonymous
  • anonymous
oops sorry I meat 5184 not 51840?
anonymous
  • anonymous
I got...soooo sorry it's 2am here and I'm tired... Thanks again for the help
kirbykirby
  • kirbykirby
np
anonymous
  • anonymous
how can I know my answer is correct?
kirbykirby
  • kirbykirby
um check it with your prof? There isn't really a way to check your answer, other than being confident in your solution

Looking for something else?

Not the answer you are looking for? Search for more explanations.