One tank contains 40 gallons of pure water. Brine with 3 lb / gal of flute salt in the size of 2 gal / min and the mixture escapes at a rate of 3 gal / min.A.-Calculate the amount of salt in the tank when it is only left in the 20 gallons.B.-How much is the maximum amount of salt in the deposit?

Answer:A- 45 lbB- 48.188 lbStep-by-step explanation:Let q(t) be the amount of salt in the tank at a given moment t. The rate of change of salt with time, q'(t), equals the difference between the speed of salt entering minus the speed of salt leaving. The speed at which the salt is entering 2 gal/min with a concentration of 3lb/gal, i.e.,  the salt is entering at 6 lb/min. On the other hand, since the volume of the tank is  40 -(2t-3t) = 40-t the salt is leaving at a speed of 3 gal/min with a concentration of q(t)/(40-t) lb/gal, so  the salt is leaving at a speed of 3q(t)/(40-t). We then have the following differential equation [tex]q'(t)=6-\frac{3q(t)}{40-t}[/tex] This is an ordinary linear differential equation of first order that can be written as [tex]q'(t)+(\frac{3}{40-t})q(t)=6[/tex] The solution of this equation is [tex]q(t)=c(t-40)^3-3t+120[/tex] where c is a constant that can be determined with the initial condition q(0)=0 [tex]q(0)=c(-40)^3+120=0\Rightarrow -64000c=-120\Rightarrow c=0.001875[/tex] and finally A- Since the volume is 40-t, when t = 20, there are left 20 gallons and  [tex]\boxed{q(20)=0.001875(20-40)^3-3*20+120=45 lb}[/tex] B- Let's find the critical points of q(t) [tex]q'(t)=0.005625(t-40)^2-3=0\Rightarrow t=16.906;\;t=63.094[/tex] but when t = 63.094 the tank is already empty, so the only possibility is t = 16.906. Let's verify this is, indeed, a maximum [tex]q''(t)=0.01125(t-40)\Rightarrow q''(16.906)<0[/tex] so, the maximum is attained at t = 19.906. At this moment, the amount of salt is [tex]\boxed{q(16.906)=0.001875(16.906-40)^3-3*16.906+120=48.188 lb}[/tex]