An Egyptian pharaoh ALGEBRA 2 Exam 2contin n Egyptian pharao

ALGEBRA 2 Exam 2-contin n Egyptian pharaoh has ordered the construction of four square pyramids. The main pyramid will be the tallest, while the other three pyramids will each be half its height. All four pyramids must have a height to base length ratio of 2:3. The pharaoh has allocated a maximum amount of 30 million cubic feet of stone for construction. Find the largest possible values for the base length and height, to the nearest foot, as well as the volume of each pyramid, in 29. scientific notation to 3 decimal places. Hint: Recall that the volume of a pyramid is vB, where B is the area of the base and h is the height.

Solution

Let the height and the base of the main pyramid be h and b respectively and let h₁ and b₁ be the height and the base of the other 3 pyramids. Since the height to base ratio of each pyramid is 2: 3, we have h/b = 2/3 or, h = 2b/3 ( or, b = 3h/2) and h₁ / b₁ = 2/3 or, h₁ = 2b₁ /3. Further, since the height of the 3 smaller pyramids is half that of the main pyramid, we have h₁ = h/2 so that h/2 = 2b₁/3 or b₁ = 3h/4. The volume of the main pyramid is v = (1/3)bh = ( 1/3) h(3h/2) = h²/2. Similarly, the volume of each of the other pyramids is v₁ = (1/3)b₁h₁ = (1/3) (3h/4)( h/2)= h²/8. Thus the volume of all the four pyramids together is h²/2 + 3h²/8 = 7h²/8. Now, since the maximum quantity of stone available is 30 million cubic feet, we have 7h² / 8 = 30 or, h² = 30*8/7 = 240/7 = 34.28571429 so that the maximum height of the main pyramid is h = 34.28571429 = 5.855400438 million feet = 5855400.438 feet. Its maximum base is 3h/2 = (3/2)* 5855400.438 = 8783100.657 feet. Further, themaximum height of each of the other 3 pyramids is h/2 =(1/2)* 5855400.438 = 2927700.219ft. Also, the maximum base of each of the smaller pyramids is 3h/4 =(3/4)* 5855400.438 = 4391550.329 feet.