Results 1 to 2 of 2

Math Help - Great Pyramid question...need help!!!!!

  1. #1
    Newbie
    Joined
    Jun 2008
    Posts
    1

    Unhappy Great Pyramid question...need help!!!!!

    this is high school calculus
    and i have no idea how to do this

    question:
    It is reported that the Great Pyramid of Egypt was built in 20 years.
    A side of its square has measures 755 feet, and its height is 410 feet.
    The stone making up the Pyrimid has a density of approximately 200 pounds per cubic foot.
    Assume that a typical worker lifted ten 50-pounds blocks a distance of 4 feet every hour and that he worked 10 hours a day,
    300 days a year. Estimate how many workers were needed to build the Great Pyrimid of Egypt.

    the answer is around xx,000 people


    thx!
    Last edited by sketch62989; June 1st 2008 at 10:05 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Jun 2008
    Posts
    2

    Phew!

    Ok here's how I got it...

    First, find out how the size of the base varies with height. Draw a triangle from the corner of the base to the centre of the base, then up to the peak. This lets you find the angle of the corners, about 56.9 deg. Remember that the distance to the centre is 0.5*sqrt2*the length of side

    The length of the side then becomes

    (2*sqrt2/tan56.9)*distance from the top of the pyramid
    =1.84 * distance from the top. (we'll call this "t")
    =1.84 * t

    Now find the volume of the pyramid that starts "t" from the peak

    V=1/3 * A * t
    subbing in our length of side and squaring it to find A
    V = 1.133*t^3

    Find the mass of the pyramid that's "t" from the peak
    M = 200lbs * V
    = 226.6 * t^3 lbs*cubic ft

    So now we have to add up all the mass that has to be lifted upwards. As we're summing, that's an integration problem

    Also, now we have to deal with "height off the ground" as we're lifting. So into our last equation we'll sub "410-r" where r is the height off the ground.

    It's a definite integral from 0 to 410.

    INT (226.6 * (410-r)^3)dr

    I think you have to expand out the cubic term before you integrate and you can take the constant to before the integral.

    226.6 * INT (6.89*10^7 - 5*10^5*r + 1230*r^2 - r^3) dr

    = 226.6 * (6.89*10^7 *r - 2.5*10^5*t^2 + 410r^3 - (1/4)*t^4) evaluated between 0 and 410

    we can just sub in 410 for "r" as when r=0, the integral=0
    so when we work it all out we get
    Total stone to be lifted = 1.12*10^13 lbs.ft

    One worker = 10blocks*50lbs*4ft*10hours*300days*20 years
    = 1.2*10^8 lbs.ft

    Divide stone by stone per worker = 93,333 workers = THE END

    I would strongly recommend going back over all the equations because I rounded heavily in places to avoid having to write too much. But I think that's how you do it. Good luck!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. problem in pyramid question
    Posted in the Geometry Forum
    Replies: 1
    Last Post: March 30th 2010, 03:36 AM
  2. Pyramid Frustum Question
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 22nd 2010, 11:12 AM
  3. Hypergeometric/Paslow's Pyramid Question?
    Posted in the Statistics Forum
    Replies: 3
    Last Post: November 8th 2009, 03:37 PM
  4. Great card game question: Pitch
    Posted in the Statistics Forum
    Replies: 1
    Last Post: January 14th 2008, 03:45 PM
  5. Question on bounds - volume of pyramid
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: February 27th 2007, 03:48 PM

Search Tags


/mathhelpforum @mathhelpforum