Reconstructing a Day's Output from Pulp Meter Readings
A cold-drink unit in Lucknow blends three fruit drinks, P, Q and R, from mango, apple and banana pulp. Each litre of P uses 0.5 L mango, 0.3 L apple and 0.2 L banana pulp; each litre of Q uses 0.2 L mango, 0.5 L apple and 0.3 L banana; each litre of R uses 0.3 L mango, 0.2 L apple and 0.5 L banana. On Monday the plant drew 48 L mango, 49 L apple and 53 L banana pulp, and the accountant has to enter the day's output of each drink in the stock register. On Tuesday the dip readings were 40 L mango, 60 L apple and 20 L banana, and the accountant is asked to sign the register before the pulp tanks are refilled. The plant manager insists the meters are reliable because the tanks are dipped by hand twice a day, and warns that any query will delay the month's excise filing.
1. Set up Monday's figures as a matrix equation, compute the determinant of the coefficient matrix, and state precisely what the determinant does and does not tell the accountant.
Read the worked answer
Let p, q, r be the litres of P, Q, R produced. Reading column by column from the recipes:
0.5p + 0.2q + 0.3r = 48 0.3p + 0.5q + 0.2r = 49 0.2p + 0.3q + 0.5r = 53
M = [[0.5, 0.2, 0.3], [0.3, 0.5, 0.2], [0.2, 0.3, 0.5]].
det M = 0.5(0.25 − 0.06) − 0.2(0.15 − 0.04) + 0.3(0.09 − 0.10) = 0.095 − 0.022 − 0.003 = 0.07 ≠ 0.
Because det M ≠ 0, M is invertible and the system has exactly one solution: the readings correspond to one and only one production triple. What this does not prove is that the triple is physically possible - a unique solution can still contain a negative number of litres. Uniqueness and plausibility are two separate tests, and Tuesday's figures will need the second one.
2. Solve Monday's system and verify the answer against the banana row.
Read the worked answer
The cofactor matrix of M is [[0.19, −0.11, −0.01], [−0.01, 0.19, −0.11], [−0.11, −0.01, 0.19]], so adj M (its transpose) is [[0.19, −0.01, −0.11], [−0.11, 0.19, −0.01], [−0.01, −0.11, 0.19]] and M⁻¹ = (adj M)/0.07.
p = (0.19·48 − 0.01·49 − 0.11·53)/0.07 = (9.12 − 0.49 − 5.83)/0.07 = 2.80/0.07 = 40 q = (−0.11·48 + 0.19·49 − 0.01·53)/0.07 = (−5.28 + 9.31 − 0.53)/0.07 = 3.50/0.07 = 50 r = (−0.01·48 − 0.11·49 + 0.19·53)/0.07 = (−0.48 − 5.39 + 10.07)/0.07 = 4.20/0.07 = 60
Check on the unused row: 0.2(40) + 0.3(50) + 0.5(60) = 8 + 15 + 30 = 53, matching the banana reading. Monday: 40 L of P, 50 L of Q, 60 L of R.
3. Apply the same inverse to Tuesday's readings. What should the accountant enter, and how should the manager's insistence be answered?
Read the worked answer
With d = (40, 60, 20):
p = (0.19·40 − 0.01·60 − 0.11·20)/0.07 = (7.60 − 0.60 − 2.20)/0.07 = 4.80/0.07 = 68.6 q = (−0.11·40 + 0.19·60 − 0.01·20)/0.07 = (−4.40 + 11.40 − 0.20)/0.07 = 6.80/0.07 = 97.1 r = (−0.01·40 − 0.11·60 + 0.19·20)/0.07 = (−0.40 − 6.60 + 3.80)/0.07 = −3.20/0.07 = −45.7
The equations are solvable (det M ≠ 0), but the solution says the plant produced −45.7 litres of R, which is impossible. The fault therefore lies in the data, not in the model: at least one of Tuesday's three readings is wrong. The banana figure is the obvious suspect, because r is most sensitive to it - raising the banana reading by Δ raises r by 0.19Δ/0.07 = 2.71Δ, so the banana reading would have to be at least 20 + 16.8 ≈ 36.8 L for r to become non-negative.
The accountant should not sign, and should not silently round the negative to zero. Answer the manager by noting that a non-zero determinant guarantees a unique answer but cannot guarantee a sensible one; the sign check is the audit step, and the tanks should be re-dipped before refilling.