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I have the volume of the cone.how much cany can go in cone?
- Thread starter stern4488
- Start date
arthur ohlsten
Full Member
- Joined
- Feb 20, 2005
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- 852
A simple solution would be given by," how many volumes of [2x1x.5=1 cm^3] can fit into a volume of 10491.53 cm^3?
10491 pieces
I hope this is correct
Arthur
10491 pieces
I hope this is correct
Arthur
Arthur, that would assume you melted the candies to liquid....right?
Question makes little sense. And why pick volume of 10491.53 ?
If a square box 22by22by22 = 10648 was used instead, then it would make more sense.
Where did you get that problem, Stern? What grade are you in?
Question makes little sense. And why pick volume of 10491.53 ?
If a square box 22by22by22 = 10648 was used instead, then it would make more sense.
Where did you get that problem, Stern? What grade are you in?
Actually, melting the candies would yield 10491.53 pieces, perfectly. We make the assumption that no piece can be split, and thus at most 10491 pieces can fit. It's a neat problem I remember in grade school. Sometimes we just have to make assumptions on these. (perhaps the cone is so flat the .53 cm[sup:mucasqlu]3[/sup:mucasqlu] is air between?)
I understand the quarum of actual pieces being square and the shape they fit into being a cone. They are not going to fit exactly and it will be less than 10491, but without knowing angles of the cone or a radius we will not get a nice function to plug into an integral, no? My calculus is a wee bit rusty, but it may be beyond the scope of this part of the forum.
I understand the quarum of actual pieces being square and the shape they fit into being a cone. They are not going to fit exactly and it will be less than 10491, but without knowing angles of the cone or a radius we will not get a nice function to plug into an integral, no? My calculus is a wee bit rusty, but it may be beyond the scope of this part of the forum.
helsfire said:Actually, melting the candies would yield 10491.53 pieces, perfectly. We make the assumption that no piece can be split, and thus at most 10491 pieces can fit. It's a neat problem I remember in grade school. Sometimes we just have to make assumptions on these. (perhaps the cone is so flat the .53 cm[sup:lgkpr8es]3[/sup:lgkpr8es] is air between?)
I understand the quarum of actual pieces being square and the shape they fit into being a cone. They are not going to fit exactly and it will be less than 10491, but without knowing angles of the cone or a radius we will not get a nice function to plug into an integral, no? My calculus is a wee bit rusty, but it may be beyond the scope of this part of the forum.
Well beyond. Packing problems like this belong in the "Advanced Math" section.
My guess is he's trying to win a contest.