Pollutant concentration and the valid inverse model
A water-treatment tank is monitored for 12 hours after a filter is switched on. The pollutant concentration, C mg/L, is modelled by C(t) = (3t + 20)/(t + 4), where t is the time in hours, 0 ≤ t ≤ 12. The plant's operating rule says that the water may be released once C ≤ 3.8 mg/L; before that it must be recirculated. At t = 0 a supervisor records C = 5.0 mg/L and claims that the concentration will fall below 3.8 mg/L 'well before six hours'. A second operator needs to set a control display that shows time remaining to reach a target concentration, so the inverse function is required. The display can accept inputs only over the concentrations actually observed during the 12-hour run. There is also a proposal to extend monitoring beyond 12 hours, but no data are available to justify the same model. Use the model exactly as stated and give numerical answers to three significant figures where appropriate.
1. Find C^{-1}(y) and state its domain and range for the 12-hour monitoring period. Explain why the unrestricted algebraic inverse cannot be used for the control display.
Read the worked answer
Let y = C(t). Then y = (3t + 20)/(t + 4). Since t ≥ 0, t + 4 > 0, so multiplication is safe: y(t + 4) = 3t + 20, hence yt + 4y = 3t + 20, so t(y − 3) = 20 − 4y and t = (20 − 4y)/(y − 3). On 0 ≤ t ≤ 12, C′(t) = [3(t + 4) − (3t + 20)]/(t + 4)^2 = −8/(t + 4)^2 < 0, so C decreases. C(0) = 5.00 and C(12) = 56/16 = 3.50. Therefore C^{-1}(y) = (20 − 4y)/(y − 3), with domain 3.50 ≤ y ≤ 5.00 and range 0 ≤ t ≤ 12. The algebraic formula alone has domain y ≠ 3 and would give negative times for y > 5, so it cannot be used outside the observed interval.
2. Solve C(t) ≤ 3.8 and assess the supervisor's claim. Include the exact time threshold.
Read the worked answer
Since t + 4 > 0, C(t) ≤ 3.8 gives 3t + 20 ≤ 3.8(t + 4) = 3.8t + 15.2. Thus 3t − 3.8t ≤ 15.2 − 20, i.e. −0.8t ≤ −4.8, so t ≥ 6.00. At t = 6, C = 38/10 = 3.80 exactly. Therefore the water may be released at or after 6.00 h, not before. The supervisor's 'well before six hours' is false; the threshold is reached exactly at six hours under the model.
3. A colleague says that because C tends to 3 mg/L as t becomes large, the display can safely accept any target above 3 mg/L. Comment.
Read the worked answer
Algebraically, C = (3t + 20)/(t + 4) → 3 as t → ∞. However, the model was specified only for 0 ≤ t ≤ 12. Extrapolating beyond 12 h assumes the same filter behaviour and no change in inflow or removal; no data support that. The limiting value 3 mg/L is an algebraic feature, not evidence that release will occur at a particular later time. The display should not accept values below 3.50 without recalibration or new data.