Calculus II Problem-How many laborers would have been required to construct the pyramid?

The Problem: Suppose that it took 20 years to construct the great pyramid of Khufu at Gizeh in Egypt. This pyramid is 500 feet high and has a square base with edge length 750 feet. Suppose also that the pyramid is made of rock with density ρ = 120 lb/cubic foot. Finally suppose that each laborer did 160 ft/lb/hour of work in lifting rocks from ground level to their final position in the pyramid and worked 12 hours daily for 330 day/year. How many laborers would have been required to construct the pyramid?|||In order to do this, assume that the only distance that each laborer has to do is carry the rocky material up to the required spot in the pyramid.





Consider the amount of work that is required to move a slice of pyramid of height ∆x x units up to it's required spot. First, note that the weight of the block is:


Weight = dV = 120V,





but we don't know the volume of the block. To find the volume of the slice, note that the cross-sections of a square pyramid are squares. To find the length of the sides of the particular square cross-section at a height of x, we need to use similar triangles. Create two triangles: one by dropping a line from the apex to the center of the base and drawing a line from the center to the base to the midpoint of one side, and one by dropping a line from the apex to the middle of the cross-section and connecting the center to the side of the pyramid (this is a lot easier to see if you draw this out). These two triangles are similar, so:


(w/2)/(5 - x) = (750/2)/500 ==%26gt; w = 15x/2.





Thus, the area of the cross-section is:


A = s^2 = (15x/2)^2 = 225x^2/4,





and it's volume is 225x^2/4 * ∆x.





The weight of this cross-section is:


weight = density * volume


= 120(225x^2/4)∆x


= 6750x^2∆x.





This is the required force to move this slice to the required place. The block needs to be moved x feet, so:


dW = Fd = 6750x^3∆x.





Integrating from 0 to 500 gives the required force to construct the pyramid to be:


W = ∫ 6750x^3 dx (from x=0 to 500) = 105,468,750,000,000 ft-lbs.





Now, each worker can do 160 ft-lb per hour, so he can do 160*12*330 = 633,600 ft-lb of work per year, so 105,468,750,000,000/633,600 ≈ 166,459,517 laborers would have been required.





I hope this helps!|||This was wrong the integral was...120(750-1.5y)^2y dy from 0 to 500. But thanks for trying.

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