Production plan with a supplier-dependent oil coefficient
Alimentación Rías is a small cooperative in Galicia that prepares three weekly meal batches: A, B and C. Each batch of A uses 1 hour of kitchen time, 2 kg of potatoes and 1 litre of oil; each B uses 1 hour, 1 kg and 2 litres; each C uses 1 hour, 0 kg and t litres, where t is set by a new oil supplier. The cooperative has 60 kitchen hours, 100 kg of potatoes and 80 litres of oil per week. Let x, y and z be the numbers of batches of A, B and C. The manager writes the system x+y+z=60, 2x+y=100, x+2y+tz=80. She must decide whether the production plan is unique when t=1, and what happens when the supplier sets t=3. She also wants a feasible plan with all three meals produced when t=3, because the cooperative has already advertised the full range.
1. Compute the determinant of the coefficient matrix as a function of t. For which values of t does the system have a unique solution? Explain what the determinant test does and does not tell the manager.
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The coefficient matrix is A = [[1,1,1],[2,1,0],[1,2,t]]. Expanding along the first row: det(A) = 1*(1t - 02) - 1*(2t - 01) + 1*(22 - 11) = t - 2t + 3 = 3 - t. Therefore det(A) is nonzero exactly when t ≠ 3, and then Cramer's rule gives a unique solution. At t=3 the determinant is zero, so the system is either inconsistent or has infinitely many solutions; the determinant alone does not distinguish those cases, so row reduction is needed. The manager should not conclude that t=3 means no plan exists.
2. Solve the system for t=1. Can the cooperative produce all three meals under that plan? Justify.
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At t=1 the equations are x+y+z=60, 2x+y=100, x+2y+z=80. Subtracting the first equation from the third gives y=20. Then 2x+20=100, so x=40. The first equation gives 40+20+z=60, so z=0. The unique plan is (40,20,0). It does not produce all three meals: product C is zero. If the cooperative has promised the full range, t=1 is not acceptable, even though the system has a unique solution.
3. For t=3, solve the system and give one feasible plan with x, y and z all positive. If the cooperative wants z ≤ 10, what range of z is feasible?
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At t=3 the third equation is x+2y+3z=80. Subtract the first equation: y+2z=20. From 2x+y=100 we get y=100-2x. Substituting into y+2z=20 gives 100-2x+2z=20, so x-z=40, or x=40+z. Then y=20-2z, and z is free. Non-negativity requires z ≥ 0 and y ≥ 0, so z ≤ 10; x ≥ 0 is automatic. Thus there are infinitely many feasible plans with 0 ≤ z ≤ 10. All three positive requires 0 < z < 10; for example z=5 gives (x,y,z)=(45,10,5). If the manager imposes z ≤ 10, the same interval applies, but z=10 gives y=0, so it is not a full-range plan.