The amounts x 1 (t) and x 2 (t) of salt in two brine tanks satisfy the following

Question

The amounts x1(t) and x2(t) of salt in two brine tanks satisfy the following differential equations:

dx1/dt = -k1x1, dx2/dt = k1x1 - k2v2,

where ki = r/Vi. Solve for x1(t) and x2(t), assuming that r = 10 (gal/min), x1(0) = 15 (lb), and x2(0) = 0. Then find the maximum amount of salt ever in tank 2.

V1 = 50(gal), V2 = 25(gal)

I've solved the first part, but just can't figure out how to solve for the maximum amount of salt in tank 2. For the first part I got:

x1(t) = 15e-.2t
x2(t) = 15(e-.2t - e-.4t)

Both of those are correct based on the back of the book. For the second part, they claim that x2(5ln(2)) = 3.75 lb. But, as I've stated, I have no idea how they figured it out.

I've tried taking the derivative of x2(t) and setting it to zero:

0 = 6e-.4t - 3e-.2t

And end up with e-.2t/e-.4t = 2, but I have no idea how the 5 would come about by taking the natural log of both sides. Step by step solution would be greatly appreciated.

Explanation / Answer

DX1/DT=-K1X1

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