Answer:
A. 171.24 Ibs
Explanation:
To find the amount of salt in the tank,
Let Q = Amount of salt in the mixture
And let 100 + (3-2)t = 100 + t be the volume of mixture at anytime t.
Rate of gain - Rate of loss = dQ / dt
Concentration of salt = Q / (100+t)
For the linear differential equation,
dQ / dt = 3(2) - 2 [Q/ (100 + t)]
dQ /dt + Q [2 / (100 + t)] = 6
The general solution of the linear differential equation is:
Q (i.f) = ∫ A(t) (i.f) dt + C
Therefore,
i.f = e ^ ∫ P(t) dt
And P(t) = 2 / (100 + t)
i.f = e ^ ∫ 2 / (100 + t)
= e ^ 2㏑ (100 + t)
= e ^ ㏑ (100 + t) ^2 = (100 + t) ^2
Q(100 + t) ^ 2 = ∫6 (100 + t) ^2 dt + C
Q(100 + t) ^2 = 2(100 + t) ^ 3 + C
When t = 0, Q = 50
Therefore,
50( 100) ^2 = 2(100) ^3 + C
C = -1.5 * 10 ^6
therefore, when t = 30,
Q (100 + 30) ^2 = 2(100 + 30) ^3 - 1.5 * 10 ^6
Q (400) ^2 = 2(130) ^3 - 1.5 * 10 ^6
Q = 171.24 Ibs
