A 50-gal tank initially contains 20 gal of pure water. Salt- water solution containing 0.5 lb of salt for each gallon of water begins entering the tank at a rate of 4 gal/min. Simultaneously, a drain is opened at the bottom of the tank, allowing the salt-water solution to leave the tank at a rate of 2 gal/min. What is the salt content (lb) in the tank at the precise moment that the tank is full of salt-water solution?

x(15) = 21 lb

Step-by-step explanation:

Rate of change in volume of salt water solution = rate of volume incoming - rate of volume outgoing

                                dV/dt  = 4 - 2 =2gal/min

So, the equation for volume at cetain time t at given conditions and values becomes,

                                V(t) = 2t + V

                                V(t) = 2t + 20   galeuqation (1)

Rate of change in amount of salt = rate of salt in - rate of salt out

                                  dx/dt = {0.5*4} - {[x(t)/V(t)]*2}

                                  dx/dt = 2-2[(x(t))/(2t+20)]

                                  dx/dt = 2-[(x(t))/(v(t))] lb/min

Now, with integrating factor, we get

          exp[∫(1/(1+10))dt)] = t+10

the equation becomes

(t + 10)*x' + x = 2*(t+10)

((t+10)*x') = 2*(t+10)

(t+10)*x = t² + 20t + C

As x(0) = 0,

x(t) = (t²+20t)/(t+10)

x(15) = (15²+20*15)/(15+10)

x(15) = 21 lb

i think it would be 42cm