A 1000-liter (L) tank contains 500 L of water with a salt concentration of 10 g/L. Water with a salt concentration of 50 g/L flows into the tank at a rate of Rin=80 L/minutes (min). The fluid mixes instantaneously and is pumped out at a specified rate Rout. Let y(t) denote the quantity of salt in the tank at time t. Assume that Rout=40L/min. (a) Set up and solve the differential equation for y(t). (b) What is the salt concentration when the tank overflows?

Answer:a) [tex]y(t)=50000-49990e^{\frac{-2t}{25}}[/tex]b) [tex]31690.7 g/L[/tex]Step-by-step explanation:By definition, we have that the change rate of salt in the tank is [tex]\frac{dy}{dt}=R_{i}-R_{o}[/tex], where [tex]R_{i}[/tex] is the rate of salt entering and [tex]R_{o}[/tex] is the rate of salt going outside.Then we have, [tex]R_{i}=80\frac{L}{min}*50\frac{g}{L}=4000\frac{g}{min}[/tex], and[tex]R_{o}=40\frac{L}{min}*\frac{y}{500} \frac{g}{L}=\frac{2y}{25}\frac{g}{min}[/tex]So we obtain.  [tex]\frac{dy}{dt}=4000-\frac{2y}{25}[/tex], then[tex]\frac{dy}{dt}+\frac{2y}{25}=4000 [/tex], and using the integrating factor [tex]e^{\int {\frac{2}{25}} \, dt=e^{\frac{2t}{25}[/tex], therefore  [tex](\frac{dy }{dt}+\frac{2y}{25}}=4000)e^{\frac{2t}{25}[/tex], we get   [tex]\frac{d}{dt}(y*e^{\frac{2t}{25}})= 4000 e^{\frac{2t}{25}[/tex], after integrating both sides [tex]y*e^{\frac{2t}{25}}= 50000 e^{\frac{2t}{25}}+C[/tex], therefore [tex]y(t)= 50000 +Ce^{\frac{-2t}{25}}[/tex], to find [tex]C[/tex] we know that the tank initially contains a salt concentration of 10 g/L, that means the initial conditions [tex]y(0)=10[/tex], so [tex]10= 50000+Ce^{\frac{-0*2}{25}}[/tex][tex]10=50000+C\\C=10-50000=-49990[/tex]Finally we can write an expression for the amount of salt in the tank at any time t, it is [tex]y(t)=50000-49990e^{\frac{-2t}{25}}[/tex]b) The tank will overflow due Rin>Rout, at a rate of [tex]80 L/min-40L/min=40L/min[/tex], due we have 500 L to overflow [tex]\frac{500L}{40L/min} =\frac{25}{2} min=t[/tex], so we can evualuate the expression of a) [tex]y(25/2)=50000-49990e^{\frac{-2}{25}\frac{25}{2}}=50000-49990e^{-1}=31690.7[/tex], is the salt concentration when the tank overflows