Furniture Workshop: Singular Matrix and Infeasible Production Plan
A small furniture unit makes three models: stools, chairs and tables. Each stool uses 1 unit of timber, 2 units of labour and 3 units of varnish; each chair uses 2, 5 and 8 units; each table uses 3, 8 and k units, where k is the varnish required per table after a supplier changed the formulation. This week the workshop has 14 units of timber, 33 units of labour and 50 units of varnish. Let x, y and z be the numbers of stools, chairs and tables produced. The planner writes the system x + 2y + 3z = 14, 2x + 5y + 8z = 33, 3x + 8y + kz = 50, and asks whether a unique production plan exists. The manager will accept only non-negative whole-number quantities. A second consignment could raise the varnish stock to 52 units while k remains 13. The planner needs a clear recommendation for both the 50-unit and 52-unit varnish cases, not just an algebraic solution.
1. For the 50-unit varnish stock, find the value of k for which the coefficient matrix is singular, and determine whether the system is consistent at that value. Show the determinant and the elimination.
Read the worked answer
The determinant is D = |1 2 3; 2 5 8; 3 8 k|. Expanding along the first row:
D = 1(5k − 64) − 2(2k − 24) + 3(16 − 15) = 5k − 64 − 4k + 48 + 3 = k − 13.
So D = 0 exactly when k = 13. Put k = 13 and use the first two equations. Subtract twice the first from the second:
(2x + 5y + 8z) − 2(x + 2y + 3z) = 33 − 28, so y + 2z = 5, hence y = 5 − 2z.
The first equation then gives x = 14 − 2(5 − 2z) − 3z = 4 + z. Substitute into the third equation with k = 13:
3(4 + z) + 8(5 − 2z) + 13z = 12 + 3z + 40 − 16z + 13z = 52.
But the varnish stock is 50, not 52. Since 52 ≠ 50, the system is inconsistent: there is no solution at all, not infinitely many.
2. For k = 14, solve the system and decide whether the solution is acceptable to the manager. Explain the decision.
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With k = 14, D = 1 ≠ 0, so there is a unique algebraic solution. Use the same elimination. From the first two equations, as before:
y + 2z = 5.
Subtract three times the first equation from the third:
(3x + 8y + 14z) − 3(x + 2y + 3z) = 50 − 42, so 2y + 5z = 8.
Substitute y = 5 − 2z:
2(5 − 2z) + 5z = 8, 10 − 4z + 5z = 8, z = −2.
Then y = 5 − 2(−2) = 9 and x = 14 − 2(9) − 3(−2) = 2. The algebraic solution is (x, y, z) = (2, 9, −2). It is not acceptable: a negative number of tables cannot be produced. The system has a unique solution, but the manager must reject this plan because it is not feasible in non-negative whole numbers.
3. For k = 13 and varnish stock 52 units, describe every non-negative whole-number production plan and recommend one.
Read the worked answer
With k = 13 and varnish stock 52, the first two equations still give y = 5 − 2z and x = 4 + z. The third equation becomes
3(4 + z) + 8(5 − 2z) + 13z = 52,
which reduces to the identity 52 = 52. Thus there is a one-parameter family of real solutions:
(x, y, z) = (4 + z, 5 − 2z, z).
Non-negativity requires z ≥ 0 and 5 − 2z ≥ 0, so 0 ≤ z ≤ 2.5. Whole-number z gives z = 0, 1, 2, hence the plans
(4, 5, 0), (5, 3, 1), (6, 1, 2).
Each uses exactly 14 timber, 33 labour and 52 varnish. Recommend (6, 1, 2) if the unit wants the maximum number of tables, or (4, 5, 0) if chairs are the priority. All three are feasible; the earlier 50-unit order has no feasible plan.