Half the volume of a 12 foot high cone-shaped pile is grade A ore while the other half is grade B ore. The pile is worth $62. One-third of the volume of a similarly shaped 18 foot pile is grade A ore while the other two-thirds is grade B ore. The second pile is worth $162. Two-thirds of the volume of a similarly shaped 24 foot pile is grade A ore while the other one-third is grade B ore. What is the value in dollars ($) of the 24 foot pile?
Answers
Answer:
Let's first find the volumes of the three piles:
The first pile has a height of 12 feet, and since it's a cone, we can use the formula V = (1/3) πr^2h to find its volume.
Step-by-step explanation:
Since we know that half of the volume is grade A ore and half is grade B ore, we can divide the volume by 2. Using r as the radius, we get:
V1/2 = (1/3)πr^2(12) / 2 = (1/6)πr^2(12)
The second pile has a height of 18 feet, and we can use the same formula to find its volume. Since one-third of the volume is grade A ore and two-thirds is grade B ore, we can multiply the volume by 1/3 to get the volume of the grade A ore. Using r as the radius, we get:
V1/3 = (1/3)πr^2(18) / 3 = (1/9)πr^2(18)
The third pile has a height of 24 feet, and we can use the same formula again. Since two-thirds of the volume is grade A ore and one-third is grade B ore, we can multiply the volume by 2/3 to get the volume of the grade A ore. Using r as the radius, we get:
V2/3 = (1/3)πr^2(24) / 3 * 2 = (1/18)πr^2(24)
Now, let's use the given information about the values of the piles to form equations and solve for the value of the third pile:
The first pile is worth $62, so the value of the grade A ore is also $62. This means:
62 = (1/2)(V1/2)(value per unit volume)
62 = (1/2)((1/6)πr^2(12))(value per unit volume)
value per unit volume = 62 / ((1/2)((1/6)πr^2(12)))
value per unit volume = 372 / πr^2
The second pile is worth $162, so the value of the grade A ore is (1/3)($162) = $54. This means:
54 = (1/3)(V1/3)(value per unit volume)
54 = (1/3)((1/9)πr^2(18))(value per unit volume)
value per unit volume = 54 / ((1/3)((1/9)πr^2(18)))
value per unit volume = 162 / πr^2
Now we can equate the value per unit volume of the two piles to get an equation in terms of r:
372 / πr^2 = 162 / πr^2
210 / πr^2 = 1
r^2 = 210 / π
r ≈ 8.15 feet
Finally, we can use the volume formula and the value per unit volume of the third pile to find its value:
V2/3 = (1/3)π(8.15)^2(24) / 3 * 2 ≈ 2,018.7 cubic feet
value of pile = (2/3)(V2/3)(value per unit volume)
value of pile = (2/3)(2,018.7)(210 / π)
value of pile ≈ $8,356.60
Therefore, the value of the 24-foot pile is approximately $8,356.60.
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