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anonymous

  • 5 years ago

I need help with this calculus question. Some ancient Egyptians decided to build a pyramid with a square base that measured 750 feet on a side. The height of the pyramid was 500 feet. The density of the rock used in construction was 120 pounds per cubic foot. It was assumed that the work force would remain constant. Each laborer could do 160 foot pounds of work per hour, working 12 hour days for 330 days per year for 20 years. Determine, under these conditions, the number of laborers needed to construct the pyramid.

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  1. amistre64
    • 5 years ago
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    Is it a solid pyramid? you know, for ceremonial purposes :)

  2. anonymous
    • 5 years ago
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    (1/3)*(750)^2*500 ft^3 * 120lbs/ft^3 * 1hr/(160lbs)/12hrs

  3. anonymous
    • 5 years ago
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    yeah its solid

  4. anonymous
    • 5 years ago
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    etc

  5. anonymous
    • 5 years ago
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    i know its a definite integral equation that is the density * volume * distance(or ht.) I just can't figure out how to represent the volume.

  6. anonymous
    • 5 years ago
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    888 laborers

  7. amistre64
    • 5 years ago
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    (S) A(y) dy {0,h} or something like that.... but the formula for any conelike volume is (1/3)(Base Area)(height)...

  8. amistre64
    • 5 years ago
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    intuition would say, find the amount of rock you need, figure out how much one laborer does in the alloted time span, then multiply that by how many laborers it take to do the job..... but other than that, I dont know...

  9. anonymous
    • 5 years ago
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    \[\int\limits_{0}^{50}(120)(volume)(y)\] the equation to find the work needed is something like this, i just can't figure out the section for volume.

  10. anonymous
    • 5 years ago
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    50 should be 500

  11. amistre64
    • 5 years ago
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    volume is integrated thru taking sections of the base area as it moves thru the height....

  12. anonymous
    • 5 years ago
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    also the volume is representing the volume of a slice

  13. amistre64
    • 5 years ago
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    each slice is x^2 dy in volume right?

  14. amistre64
    • 5 years ago
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    and x changes with repect to height...or y

  15. amistre64
    • 5 years ago
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    what is the slope of the face?

  16. anonymous
    • 5 years ago
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    yeah basically, but the size of the slice changes since it is getting smaller going up

  17. anonymous
    • 5 years ago
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    i don't know that isn't given

  18. amistre64
    • 5 years ago
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    500/(750/2) would be a cross section of the "triangle"

  19. amistre64
    • 5 years ago
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    the slope of a face is 500/375....might need to reduce. the equation relating that to x and y is y = -500/375(x) + 500

  20. amistre64
    • 5 years ago
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    slope is 4/3 when reduced

  21. amistre64
    • 5 years ago
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    x increase by 2 for every 3/4 increase in y .... if that helps

  22. amistre64
    • 5 years ago
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    (3/2) (S) y dy {0,500} = (3/4) y^2 {0,500} 500^2 = 250000(3/4) = 750000/4 = 187500 that matches 500(750)/2 = 187500 for a flat area..... not sure how it helps tho :)

  23. amistre64
    • 5 years ago
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    x^2 is the area of a normal base and it changes with respect to "y" at a rate of (3/2)(y) dy ?? It looks right, but this is still shakey ground fer me :)

  24. anonymous
    • 5 years ago
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    yeah i think ive gotten it now, thanks.

  25. amistre64
    • 5 years ago
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    if I messed up anything, youlet me know :)

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