heartshapes
New member
- Joined
- Feb 21, 2008
- Messages
- 36
1) Find a can
2) Estimate cost ratios between the top, bottom and sides of can.
3) Measure the can dimensions
4) Finally, compute the optimal dimensions of the cylindrical can holding your given volume which will minimize the cost of material to produce that can. Notice we can do this without knowing the actual cost of the material to produce the can. So, you DO NOT need to know the actual cost per square centimeter of aluminum alloy 3104-H19 used to make your can. It is a constant number anyway so treat is as a constant C.
Okay so..
Using a soup can, I found the radius to be 3.75cm and the height to be 11.2cm, the volume in 480cm[sup:itza6xj7]3[/sup:itza6xj7].
I am saying that the cost to produce the top is 2 times the cost to produce the sides by area of material (aluminum)
I don't understand how to tie the cost in it.. if you don't have to know it.. (basically I get lost on number 4)
This is what I did..
\(\displaystyle A_{min}\, =\, 2{\pi}r^2\, +\, 2{\pi}rh\, \mbox{ and }\, h\, =\, \frac{480}{{\pi}r^2}\)
\(\displaystyle A_{min}\, =\, 2{\pi}r^2\, +\, 2{\pi}r\left(\frac{480}{{\pi}r^2}\right)\)
\(\displaystyle A_{min}\, =\, 2{\pi}r^2\, +\, \frac{960}{r}\)
\(\displaystyle A_{min}^{'} \, =\,4{\pi}r\, +\, \frac{960}{r^2}\)
\(\displaystyle 0\, =\, 4{\pi}r^3\, -\, 960\)
\(\displaystyle 960\4{\pi}\, =\, r^3\)
\(\displaystyle r\, =\, 4.24314\)
\(\displaystyle h\, =\, 8.48826\)
Then using a numberline I found that 4.24 is a minimum.. But I don't see how this includes the cost.
Any help would be greatly appreciated!
2) Estimate cost ratios between the top, bottom and sides of can.
3) Measure the can dimensions
4) Finally, compute the optimal dimensions of the cylindrical can holding your given volume which will minimize the cost of material to produce that can. Notice we can do this without knowing the actual cost of the material to produce the can. So, you DO NOT need to know the actual cost per square centimeter of aluminum alloy 3104-H19 used to make your can. It is a constant number anyway so treat is as a constant C.
Okay so..
Using a soup can, I found the radius to be 3.75cm and the height to be 11.2cm, the volume in 480cm[sup:itza6xj7]3[/sup:itza6xj7].
I am saying that the cost to produce the top is 2 times the cost to produce the sides by area of material (aluminum)
I don't understand how to tie the cost in it.. if you don't have to know it.. (basically I get lost on number 4)
This is what I did..
\(\displaystyle A_{min}\, =\, 2{\pi}r^2\, +\, 2{\pi}rh\, \mbox{ and }\, h\, =\, \frac{480}{{\pi}r^2}\)
\(\displaystyle A_{min}\, =\, 2{\pi}r^2\, +\, 2{\pi}r\left(\frac{480}{{\pi}r^2}\right)\)
\(\displaystyle A_{min}\, =\, 2{\pi}r^2\, +\, \frac{960}{r}\)
\(\displaystyle A_{min}^{'} \, =\,4{\pi}r\, +\, \frac{960}{r^2}\)
\(\displaystyle 0\, =\, 4{\pi}r^3\, -\, 960\)
\(\displaystyle 960\4{\pi}\, =\, r^3\)
\(\displaystyle r\, =\, 4.24314\)
\(\displaystyle h\, =\, 8.48826\)
Then using a numberline I found that 4.24 is a minimum.. But I don't see how this includes the cost.
Any help would be greatly appreciated!
