Mathematics - Calculus

Letx kg be the number of kilograms of salt in the tank.

Let t be the time elasped.

dx/dt = rate of change of number of kilograms of salt

= rate in - rate out

= (2)(5) - (6)x/[100 + (6 - 5)t]

= 10 - 6x/(100 - t)

Therefore, dx/dt= 10 - 6x/(100 - t)

dx/dt + 6x/(100 - t) = 10

Integrating factor,exp[∫6/(100 - t)]dt = exp(-6ln(100 - t)) = 1/(100 - t)6

So, 1/(100- t)6[dx/dt + 6x/(100 - t)] = 10/(100 - t)6

d/dt [x/(100- t)6] = 10/(100 - t)6

x/(100 -t)6 = 2/(100 - t)5 + C, where C is a constant

x = 2(100- t) + C(100 - t)6

Whent = 0, x = 0, C = -2 X 10-10

So,x = 2(100 - t) – (2 X 10-10)(100 - t)6

dx/dt =-2 + (1.2 X 10-9)(100 - t)5

d2x/dt2= -(6 X 10-9)(100 - t)4

Setdx/dt = 0,

(100 -t)5 = (5 X 109)/3

t =30.1173 min (4 d.p.)

d2x/dt2│t = 30.1173 min < 0

So,When t = 30.1173 mins, the amount of salt in the tank is the maximum.

Max.amount of salt

= 116.4712kg (4 d.p.)