Modeling with First Order DE’s — Question 10

Question 10 – Solution

Strategy. Solve the storage balance, obtain a necessary pump rate from the final reserve, and then verify that this candidate satisfies the full time constraint.

Step 1: Model and solve the storage balance. Input minus demand minus leakage gives $$S'=u-(2+t)-0.50S,\qquad S(0)=10,$$ with all rates in L/min. A particular solution is $2u-2t$; adding $Ae^{-t/2}$ and applying $S(0)=10$ gives $$\boxed{S(t)=2u-2t+(10-2u)e^{-t/2}}.$$ It has initial value $10$, and differentiation gives the required balance on any time interval on which the physical operation remains valid.

Step 2: Find and verify the minimum rate. Since $S(4)\ge 4$ is necessary, $$\boxed{u\ge u_*= \frac{12-10e^{-2}}{2(1-e^{-2})} =6+\frac 1{e^2-1}\approx 6.157\text{ L/min}}.$$ For this candidate, $6<u_*<7$ and $$S'(t)=-2+(u_*-5)e^{-t/2}<0\qquad(0\le t\le 4).$$ Thus $S$ decreases from $10$ to exactly $4$ L and never violates the reserve. The candidate is feasible, and every smaller rate fails the necessary final condition, proving optimality.

Step 3: Recover cumulative leakage. The scheduled demand totals $\int_0^4(2+t)\,dt=16$ L. Initial storage plus pumped volume equals final storage plus demand plus leakage, so $$\boxed{L_{\mathrm{leak}}=10+4u_*-16-4=4u_*-10 \approx 14.63\text{ L}}.$$

Step 4: Compare the incomplete model. Ignoring leakage gives $S_0=10+(u-2)t-t^2/2$. This concave function has its minimum on $[0,4]$ at an endpoint; $S_0(4)\ge 4$ gives the no-leakage minimum $u=2.5$ L/min. That rate is below $u_*$ and fails the actual reserve constraint. It may eventually exhaust the actual reservoir, so negative values of its unconstrained storage formula would not describe physical stored liquid. The optimal design above stays positive and avoids that limitation.