A tank that holds 4000 gallons of water will drain completely into a reservoir in 32 minutes. The volume of remaining water in tank after time t is given by the function:
V(t)=4000(1-t/32)^2, 0<t<32
The amount of water in reservoir depends on how much water has drained out of the tank and defined by A(V)= 14000-V. After how many minutes will the amount of water in the reservoir reach 12000 gallons?
well, it looks like you want
A(V) = 12000
14000-4000(1-t/32)^2 = 12000
4000(1-t/32)^2 = 2000
(1-t/32)^2 = 1/2
1 - t/32 = ±1/√2
t/32 = 1±1/√2
t = 32(1±1/√2)
Since the domain of V is t<32, we need
t = 32(1-1/√2) = 9.37
To find the number of minutes it takes for the amount of water in the reservoir to reach 12,000 gallons, we need to find the value of t for which A(V(t)) = 12,000.
Given V(t) = 4000(1 - t/32)^2, we can substitute this into A(V) = 14,000 - V:
A(V(t)) = 14,000 - V(t)
A(V(t)) = 14,000 - 4000(1 - t/32)^2
A(V(t)) = 14,000 - 4000(1 - t/32)(1 - t/32)
A(V(t)) = 14,000 - 4000(1 - 2t/32 + t^2/1024)
A(V(t)) = 14,000 - 4000 + 8000t/32 - 4000t^2/1024
Simplifying further:
A(V(t)) = 10,000 + 250t - 3.90625t^2
We want to find t when A(V(t)) = 12,000:
12,000 = 10,000 + 250t - 3.90625t^2
Rearranging the equation and simplifying:
3.90625t^2 - 250t + 2,000 = 0
This is now a quadratic equation. We can use the quadratic formula to solve for t:
t = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = 3.90625, b = -250, and c = 2,000.
t = (-(-250) ± √((-250)^2 - 4(3.90625)(2,000))) / (2(3.90625))
t = (250 ± √(62,500 - 31,250)) / 7.8125
t = (250 ± √31,250) / 7.8125
t = (250 ± 176.78) / 7.8125
We have two solutions, t1 and t2:
t1 = (250 + 176.78) / 7.8125 = 39.32 minutes (rounded to two decimal places)
t2 = (250 - 176.78) / 7.8125 = 12.03 minutes (rounded to two decimal places)
So, the amount of water in the reservoir will reach 12,000 gallons approximately after 12.03 minutes and 39.32 minutes.