Question

3) A 200-gallon tank is half-filled with pure water. Subsequently, a salt-water solution of 3 pounds per gallon enters the tank at 2 gallons per minute. Simul- taneously, the well-stirred solution leaves the tank at 6 gallons per minute. a) Write an initial value problem (IVP) that models this process. b) Solve your IVP. c) Over what time frame is the solution of your IVP valid? Explain. d) When is the amount of salt in the tank maximized? Give the exact answer and an approximation to the nearest minute. e) What is this maximum amount of salt? Give the exact answer and an approximation to the nearest pound

Answer 1

$\dpi{100} \small \small \small \textbf{ Let S(t) be the amount of salt in the tank at time t then}\\ \textbf{Inflow information}\hspace{2cm}\textbf{Outflow information}\\ r_{i}=2\; \; \textbf{gal/min}\hspace{3cm}r_{o}=6\; \; \textbf{gal/min}\\ c_{i}=3\; \; \textbf{lb/gal}\hspace{3cm}c_{o}=?\\ {\color{Red} \textbf{Volume}}\\ \textbf{Maximum total volume}=100 \; \textbf{gal}\\ \textbf{Initial volume}=V_{0}=100 \; \textbf{gal}\\ {\color{Red} \textbf{Formula}:}\; V(t)=V_{0}+(r_{i}-r_{o})t=100-4t\\ \textbf{Concentration}\; (c_{o}):\\ {\color{Red} \textbf{Formula}:}\; (c_{o})=\frac{S(t)}{V(t)}=\frac{S(t)}{100-4t}\\ \textbf{Initial condition}\;\; S(0)=0\\ \textbf{Solving the problem for}\; S(t)\\ \textbf{For these type of problems we are the given differential equation}\\ \frac{\mathrm{d} S}{\mathrm{d} t}=r_{i}c_{i}-r_{o}c_{o}\\ \textbf{Plugging in our information we get}\\$

$\dpi{100} \small \frac{\mathrm{d} S}{\mathrm{d} t}=2\times 3-6\times \frac{S(t)}{100-4t}\\ \frac{\mathrm{d} S}{\mathrm{d} t}=6-6\times \frac{S(t)}{100-4t}\\ \frac{\mathrm{d} S}{\mathrm{d} t}+\frac{6}{100-4t}S(t)=6 \\ \textbf{I.F}=e^{\int \frac{6}{100-4t}dt}=e^{-\frac{3}{2}\ln \left|100-4t\right|}=(100-4t)^{-\frac{3}{2}}\\ S(t)\times (100-4t)^{-\frac{3}{2}}=\int (100-4t)^{-\frac{3}{2}}\times 6dt+c\\$

$\dpi{100} \small S(t)\times (100-4t)^{-\frac{3}{2}}=6\left(\frac{t}{8\left(-t+25\right)^{\frac{3}{2}}}+\frac{-3t+50}{8\left(-t+25\right)^{\frac{3}{2}}}\right)+c\\ S(t)=6(100-4t)^{\frac{3}{2}}\left(\frac{t}{2\left(-4t+100\right)^{\frac{3}{2}}}+\frac{-3t+50}{2\left(-4t+100\right)^{\frac{3}{2}}}\right)+c(100-4t)^{\frac{3}{2}}\\ S(t)=6\left(\frac{t}{2}+\frac{-3t+50}{2}\right)+c(100-4t)^{\frac{3}{2}}\\ S(t)=6\left(\frac{-2t+50}{2}\right)+c(100-4t)^{\frac{3}{2}}\\ S(t)=150-6t+c(100-4t)^{\frac{3}{2}}\\ \textbf{Initial condition}\;\; S(0)=0\\ S(0) =0\Rightarrow 150+c(100)^{\frac{3}{2}}=0\Rightarrow c=-\frac{3}{20}\quad \left(\mathrm{Decimal:\quad }c=-0.15\right)\\ \textbf{Hence,}\\ \boxed {{\color{Red} S(t)=150-6t-0.15(100-4t)^{\frac{3}{2}}}}\\$