Modeling — Question 1

Question 1 – Solution

Strategy. Use volume flow times concentration for each salt flux, then use conservation to reduce the coupled balances.

Step 1: Construct the two balances. Each tank gains and loses $10$ L/min, so its volume is fixed. Its outgoing salt rate is $10$ times its own concentration. Thus $$\boxed{x'=-\tfrac 1{10}x+\tfrac 1{20}y,\qquad y'=\tfrac 1{10}x-\tfrac 1{20}y.}$$ The coefficients have units min$^{-1}$: equal volume flows remove different fractions of the unequal tank volumes per minute.

Step 2: Use the conserved total. Adding gives $(x+y)'=0$, so $x+y=10$ kg. At equilibrium $x/100=y/200$. Consequently $\boxed{(x_*,y_*)=(10/3,20/3)\text{ kg}}$ and the common concentration is $1/30$ kg/L. Equal amounts would leave unequal concentrations and hence a nonzero net salt transfer.

Step 3: Solve and check the physical range. Substituting $y=10-x$ gives $x'=1/2-3x/20$. The initial data yield $$\boxed{x=\tfrac{10}3+\tfrac{20}3e^{-3t/20},\qquad y=\tfrac{20}3(1-e^{-3t/20}).}$$ Their sum is $10$, both are nonnegative, and their initial values are $(10,0)$. Differentiation gives $x'=-e^{-3t/20}$ and $y'=e^{-3t/20}$ kg/min; substitution into the original flux balances gives the same rates.

Step 4: Compute a lasting tolerance time. The concentration difference is $$\frac{x}{100}-\frac{y}{200}=\frac 1{10}e^{-3t/20}>0.$$ It decreases strictly, so its first value $0.005$ occurs at $\boxed{t=\tfrac{20}3\log 20\text{ min}\approx 19.97\text{ min}}$. The inequality holds for this time and every later time. The first concentration approaches the dotted limit from above; the second approaches it from below.