Recovering Pack Counts from a Resource Matrix
A small cooperative bakery in Patiala makes two types of biscuit packs, A and B. Each pack of A uses 200 g flour and 60 g sugar; each pack of B uses 150 g flour and k g sugar. On Monday the bakery used 5100 g flour and 1800 g sugar, and the manager wants to recover how many packs of each type were made by solving a 2×2 system. The stock register also contains a proposed 'low-sugar' version of B that would use 45 g sugar per pack. A trainee sets up the matrix [[200,150],[60,k]], inverts it without checking any condition, and reports a unique answer for every k. The manager is uneasy: the two products might become indistinguishable in resource use, and a negative pack count would be nonsense. Before signing the register, the manager wants the determinant as a function of k, the value that causes trouble, and a check of the Monday figures for k = 60.
1. Compute det [[200,150],[60,k]] and identify the value of k for which the system may fail to have a unique solution. Interpret that value in bakery terms.
Read the worked answer
det = (200)(k) − (150)(60) = 200k − 9000. A 2×2 system has a unique solution exactly when this determinant is nonzero. Setting 200k−9000=0 gives k=45. At k=45, pack B uses 150 g flour and 45 g sugar. Compare the resource ratios: pack A uses 200 g flour per 60 g sugar, i.e. flour:sugar = 200:60 = 10:3; pack B uses 150:45 = 10:3. The two packs consume flour and sugar in the same proportion, so the two columns of the matrix are proportional (column 2 = 0.75 × column 1). The total usage figures then either lie in that same ratio, giving infinitely many pack combinations, or do not, giving no exact solution. The bakery could not tell from total flour and sugar alone how many of each pack were made. At k=45 a unique recovery is impossible without another measurement.
2. For k=60, solve the Monday system with total usage 5100 g flour and 1800 g sugar. Show the inverse or elimination steps and interpret the answer.
Read the worked answer
With k=60 the system is 200x+150y=5100 and 60x+60y=1800, where x and y are packs of A and B. Divide the second equation by 60: x+y=30. From this y=30−x. Substitute into the first: 200x+150(30−x)=5100, so 200x+4500−150x=5100, hence 50x=600 and x=12. Then y=30−12=18. Check: flour 200(12)+150(18)=2400+2700=5100; sugar 60(12)+60(18)=720+1080=1800. The determinant is 200(60)−150(60)=3000, nonzero, consistent with a unique solution. In bakery terms, Monday produced 12 packs of A and 18 packs of B. Both are nonnegative whole numbers, so the algebraic answer is physically admissible here.
3. The trainee says 'because the determinant is nonzero, the answer must be physically possible.' Critique this statement using the matrix and the given totals.
Read the worked answer
A nonzero determinant guarantees only a unique solution in real numbers; it does not guarantee nonnegative or integer pack counts. Keep the same matrix with k=60 and the same sugar total 1800 g, so x+y=30. If the flour total were 2000 g, then 200x+150(30−x)=2000, so 50x+4500=2000, giving x=−50 and y=80. The determinant is still 3000, and the solution is unique, but a negative number of packs is physically nonsense. Conversely, with the actual flour total 5100 g the same matrix gives x=12, y=18, which is admissible. So the trainee's inference is invalid: uniqueness is an algebraic property, while physical possibility requires an extra check that x≥0, y≥0, and ideally that x and y are integers.