Dairy Blend with a Parameter in the Recipe Matrix
Anandpur Milk Producers’ Society runs a small plant that blends three products, P1, P2 and P3, from milk powder, starter culture and labour. Each unit of P1 uses 1 kg powder, 2 L culture and 1 labour-hour; each unit of P2 uses 2 kg, 5 L and 3 hours; each unit of P3 uses 1 kg, 3 L and k labour-hours, where k is a process setting. On a particular day the society has 5 kg powder, 12 L culture and 9 labour-hours available. The manager writes the system as Ax=b, with x=(x1,x2,x3). A trainee claims the plant can always find a unique production plan because there are three products and three resources. The manager is not sure: the labour coefficient k can be changed between 1 and 5, and the society once had a day when the resources were 5 kg, 12 L and 8 labour-hours and no plan existed. The board wants a short note on when the system has a unique solution, when it has none, and what the best response is if the labour setting is fixed at k=2.
1. Write the coefficient matrix A and compute det A as a function of k. For which k in the interval [1,5] does the system have a unique solution? Does the trainee’s three-by-three argument guarantee uniqueness?
Read the worked answer
A = [[1, 2, 1], [2, 5, 3], [1, 3, k]]. Expanding along the first row:
det A = 1(5k − 9) − 2(2k − 3) + 1(6 − 5) = 5k − 9 − 4k + 6 + 1 = k − 2.
Thus det A = 0 exactly when k = 2. For k ≠ 2 the coefficient matrix is invertible, so there is a unique solution for every b. On 1 ≤ k ≤ 5, only k = 2 fails. The trainee’s argument is wrong: three equations in three unknowns guarantee neither existence nor uniqueness. The determinant or row-rank test decides the question.
2. Take k=4 and b=(5,12,9). Solve the system and state a production plan. Check the plan against the available resources.
Read the worked answer
At k=4 the system is x1 + 2x2 + x3 = 5 2x1 + 5x2 + 3x3 = 12 x1 + 3x2 + 4x3 = 9.
Subtract twice the first equation from the second: x2 + x3 = 2. Subtract the first equation from the third: x2 + 3x3 = 4. Subtracting these two results gives 2x3 = 2, so x3 = 1, then x2 = 1, then x1 = 5 − 2 − 1 = 2.
Check: row 2 gives 4 + 5 + 3 = 12; row 3 gives 2 + 3 + 4 = 9. The plan is (2,1,1): 2 units of P1, 1 unit of P2 and 1 unit of P3. Since det A = 4 − 2 = 2 ≠ 0, this is the only plan for these resources.
3. For k=2 and b=(5,12,9), show that there is no production plan. What row relation causes this, and what single change in available labour would make the system consistent? Is the resulting plan unique?
Read the worked answer
At k=2, row 2 = row 1 + row 3 because (2,5,3) = (1,2,1) + (1,3,2). Hence equation 2 is the sum of equations 1 and 3, so consistency requires b2 = b1 + b3. Here 12 ≠ 5 + 9 = 14, so the system is inconsistent: no non-negative production plan exists.
Reducing available labour from 9 to 7 makes b = (5,12,7), and then 12 = 5 + 7. The system is consistent, but the rank is 2, not 3, so there are infinitely many solutions with one free variable. The manager should not promise a unique plan at k=2; a choice rule or an extra constraint is needed to select one plan.