Dairy co-operative blending plan with an unknown culture requirement
A dairy co-operative in Erode prepares three blended products from milk, culture and sugar. Each unit of product A uses 1 litre milk, 2 kg sugar and 1 packet culture; each unit of B uses 2 litres milk, 1 kg sugar and 3 packets culture; each unit of C uses 1 litre milk, 4 kg sugar and k packets culture. On Monday the co-operative has 1,000 litres milk, 1,500 kg sugar and 1,200 packets culture, and it wants to exhaust all three inputs exactly by producing x, y and z units of A, B and C. The manager writes the system x + 2y + z = 1000, 2x + y + 4z = 1500, x + 3y + kz = 1200. Before ordering packaging, the manager must know whether the plan has a unique solution. The accountant has proposed k = 2, while the production supervisor wants k = 1/3 because it saves one packet per unit. The board will not approve a plan that implies negative production.
1. Write the system as AX = b, find det A in terms of k, and state the values of k for which the production plan is guaranteed a unique solution.
Read the worked answer
Take A = [[1, 2, 1], [2, 1, 4], [1, 3, k]], X = (x, y, z)ᵀ and b = (1000, 1500, 1200)ᵀ. Expanding det A along the first row:
det A = 1(1·k − 4·3) − 2(2·k − 4·1) + 1(2·3 − 1·1) = (k − 12) − 2(2k − 4) + 5 = k − 12 − 4k + 8 + 5 = 1 − 3k.
A unique solution exists exactly when det A ≠ 0, so k ≠ 1/3. The determinant does not involve the right-hand side, so this conclusion is about uniqueness only. At k = 1/3 the system may have no solution or infinitely many; that requires a separate consistency check.
2. For k = 2, solve the system by Cramer's rule and state whether the plan is feasible under the board's condition.
Read the worked answer
Here det A = 1 − 3(2) = −5.
Dx = det [[1000, 2, 1], [1500, 1, 4], [1200, 3, 2]] = 1000(2 − 12) − 2(3000 − 4800) + 1(4500 − 1200) = −10000 + 3600 + 3300 = −3100, so x = −3100/(−5) = 620.
Dy = det [[1, 1000, 1], [2, 1500, 4], [1, 1200, 2]] = 1(3000 − 4800) − 1000(4 − 4) + 1(2400 − 1500) = −1800 + 900 = −900, so y = −900/(−5) = 180.
Dz = det [[1, 2, 1000], [2, 1, 1500], [1, 3, 1200]] = 1(1200 − 4500) − 2(2400 − 1500) + 1000(6 − 1) = −3300 − 1800 + 5000 = −100, so z = −100/(−5) = 20.
Check: 620 + 360 + 20 = 1000; 1240 + 180 + 80 = 1500; 620 + 540 + 40 = 1200. All quantities are non-negative, so the plan is feasible: 620 units of A, 180 of B and 20 of C.
3. The supervisor prefers k = 1/3. Use row reduction on the augmented matrix to decide whether any production plan can exhaust all inputs exactly at this k, and advise the board.
Read the worked answer
At k = 1/3 the augmented matrix is [1 2 1 | 1000] [2 1 4 | 1500] [1 3 1/3 | 1200].
R2 − 2R1 gives [0 −3 2 | −500]. R3 − R1 gives [0 1 −2/3 | 200]. Multiplying the new third row by 3 gives [0 3 −2 | 600]. Adding this to the second row gives [0 0 0 | 100].
The last row reads 0 = 100, a contradiction, so the system is inconsistent. Therefore at k = 1/3 there is no production plan that uses exactly 1,000 litres of milk, 1,500 kg of sugar and 1,200 packets of culture. The supervisor's saving of one packet per unit sounds attractive, but it makes the three input targets mutually incompatible. The board should reject k = 1/3 and use a value such as k = 2, subject to the non-negativity check.