Results 1 to 8 of 8

Math Help - Simple problem

  1. #1
    Junior Member
    Joined
    Jan 2010
    Posts
    47

    Simple problem

    Hello everyone, I was completing my math project when I got stumped on the following question.

    Now, we just recently learned about polynomial functions so I'm assuming that they should be applicable to this problem, but I have yet to find out how.

    If anyone knows how or can offer some help, that would be greatly appreciated!

    Problem:

    Jasmine is a manufacturing engineer who is trying to find out if she can use a 12-inch-by-20-inch sheet of cardboard and a 15-inch-by-16-inch sheet of cardboard to make boxes with the same height and volume.

    Is such a pair of boxes possible?





    Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Aug 2007
    From
    USA
    Posts
    3,110
    Thanks
    2
    Sure, cut out a 6" x 6" square from each and throw the rest away.

    Seriously, though, you'll have to be far more specific.

    Is there a particular box design?
    Does it have a lid?
    Is it square or just rectangular?
    Are there cutting restrictions? Can we just cut the corners or can we slice and dice at will?

    Like I said, FAR more specific.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jan 2010
    Posts
    47
    I appreciate the quick help, but, there was no other information given..

    The problem (which was part of a project) was actually due today, but my group thought we had more time, so is there any way we may assume on some non-given information?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by qcom View Post
    I appreciate the quick help, but, there was no other information given..

    The problem (which was part of a project) was actually due today, but my group thought we had more time, so is there any way we may assume on some non-given information?
    this might be what they are after. we want to cut equal squares from each corner of each sheet, then we fold the flaps up to form the (open top) boxes. the side length of the squares will be the height. let this be x. then forming the volumes and equating them, we get

    V = x(20 - 2x)(12 - 2x) = x(16 - 2x)(15 - 2x)

    if there is a positive x that solves that equation, we're done. if not, it's impossible.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Jan 2010
    Posts
    47
    ok, I'm still a little confused, why do we cut it out of a square exactly?

    Also, what does x represent??

    I understand that if a positive solution is present then it's possible.
    I believe, after solving, that it isn't possible.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by qcom View Post
    ok, I'm still a little confused, why do we cut it out of a square exactly?

    Also, what does x represent??

    I understand that if a positive solution is present then it's possible.
    I believe, after solving, that it isn't possible.
    if we don't cut out squares, we couldn't form a rectangular box with this method. it would form some sort of trapezoidal prism. anyway, the idea of what i am thinking about is illustrated here

    see the diagram in post #2 in particular.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Jan 2010
    Posts
    47
    OK, thanks for the clarification, so I believe that the answer is that the pair of boxes is NOT POSSIBLE, what do you think?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by qcom View Post
    OK, thanks for the clarification, so I believe that the answer is that the pair of boxes is NOT POSSIBLE, what do you think?
    if the problem is to be interpreted as i interpreted it, then i agree.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: February 11th 2011, 03:28 AM
  2. One more simple problem
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: May 28th 2010, 02:28 PM
  3. Simple Problem
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: May 28th 2010, 01:20 PM
  4. simple problem need help
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 2nd 2009, 04:02 PM
  5. Simple problem... help please.
    Posted in the Geometry Forum
    Replies: 3
    Last Post: October 30th 2007, 02:40 AM

Search Tags


/mathhelpforum @mathhelpforum