Question

**WILL MEDAL**
A cylindrical cardboard mailing tube has a diameter of 90 mm. and a height of 600 mm. A total of 20% of the tube’s capacity is filled with spherical packing foam. Each foam sphere has a radius of 15 mm.

Approximately how many pieces of foam are in the tube?
Use 3.14 to approximate pi and express your answer as a whole number.

Answer 1

@amistre64

Answer 2

@aaronq

Answer 3

find 20% of the volume of the cylindar and divide it by the volume of a sphere is my thought

Answer 4

whats our formulas for these things?

Answer 5

My mom and I tried to find the answer and we came out to 210 or 659

Answer 6

the answer will become apparent at the end of the process; lets make sure we are using the correct stuff first

what are the formulas we will need to use? volume of cyl, and volume of sphere ...

Answer 7

Ok
volume of a sphere is 4/3*pi*r^3

Answer 8

good so far, and cylindar?

Answer 9

the volume of a cylinder = pi* r^2* h
r is the radius and h is the height

Answer 10

good, and we are going to have 2 different radius measures .. what are they?

Answer 11

3in and 4.5in?

Answer 12

20% of the cylindar divided by the volume of a sphere


.20 cylindar
----------
sphere



.20 pi R^2
------------
4/3 pi r^3

pi/pi = 1 so the factors cancel


.20 R^2
----------
4/3 r^3


.20/(4/3) = .20(3/4) = .05(3) = .15

so this simplifies to this as a formula for us:

.15 R^2/r^3

Answer 13

daims are 90 and 15 ... what are the radiuses?

Answer 14

45 and 7.5

Answer 15

good, i noticed i dropped the h someplace along the way ... h is 600

.15 h R^2/r^3

thats better

so we just fill in

.15 (600) (45)^2/(7.5)^3

Answer 16

96.426

Answer 17

hmm, you might need new batteries in your calculator :)

http://www.wolframalpha.com/input/?i=.15+%28600%29+%2845%29%5E2%2F%287.5%29%5E3

Answer 18

Um i dont think so lol I used an online one

Answer 19

So 432?

Answer 20

then your inputs were mistyped. either way ... we are getting an upper limit of 432.

yes

Answer 21

So do we multiply 432 and pi?

Answer 22

no, pi was a common factor that canceled out. there is no need to work in pi

Answer 23

Oh ok so the answer is 432

Answer 24

yes, but how good of an approximation that is i wouldnt know how to tell

we know that at most, 432 spheres is equal to the volume required .. but if you stuff them in there their shape leaves gaps ... so they fill MORE than 20%

Answer 25

Huh?

Answer 26

@amistre64

Answer 27

The answert was 54

Answer 28

how is the in any way a possibility

Answer 29

if we assume a cube instead of a sphere?

.20(3.14)(45)^2(600)/15^3 thats 226

Answer 30

did you mis type the information any?