A cylindrical tank with radius 3 meters is filled with a solution. The volume at t minutes is given by V = (-t2 + 12t + 24)Ï€ m3. At 4 minutes, there are d meters of unused vertical space in the tank. At 6 minutes, there are only 1/3d meters of space remaining. What is the height of the tank?
An alternate approach:
At the 4-minute mark, the height of the UNUSED portion of the tank -- the remaining "vertical space" -- is equal to d meters.
At the 6-minute mark, the height of the unused portion of the tank is equal to (1/3)d meters -- a decrease of 2/3.
Since the HEIGHT of the unused portion of the tank decreases by 2/3, the VOLUME of the unused portion of the tank also decreases by 2/3.
Let U = the volume of the unused portion of the tank at the 4-minute mark.
Between the 4-minute mark and the 6-minute mark, the value of U decreases by 2/3.
Volume of the solution at the 4-minute mark:
V = (-t² + 12t + 24)π = (-4² + 12*4 + 24)π = 56π cm³.
Volume of the solution at the 6-minute mark:
V = (-t² + 12t + 24)π = (-6² + 12*6 + 24)π = 60π cm³.
Between the 4-minute mark and the 6-minute mark, the volume of the solution INCREASES by 4π cm³.
As noted above, these 4π cm³ represent a 2/3 DECREASE in the value of U.
Thus:
4Ï€ = (2/3)U
U = 6π cm³.
Since the volume of the solution at the 4-minute mark is 56 cm³, and the value of U at the 4-minute mark is 6π cm³, we get:
Total volume = 56π + 6π = 62π cm³.
Since the volume of the tank is 62π cm³, and the radius is 3 cm, we get:
π(3²)h = 62π
h = 62/9.
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