Trail Mix Repackaging with Loss and Ratio Constraints
A school store is repackaging trail mix from two bulk bins. Bin X contains 24 kg of mix that is 30% nuts by mass; bin Y contains 36 kg of mix that is 55% nuts. The store wants 40 kg of a new mix that is at least 45% nuts and at most 48% nuts. It may use any amount from each bin but cannot add pure nuts or filler. The owner also insists that the amount taken from bin Y be no more than twice the amount taken from bin X, to keep the cheaper base mix in stock. A 2% loss occurs during repackaging, so only 98% of each amount taken ends up in the final 40 kg. Determine a feasible blend and check whether the nuts target can be met after loss.
1. Let x and y be the kilograms taken from bins X and Y before loss. Write the mass, nut-percentage, and stock constraints. Find the feasible range for x and give one feasible blend.
Read the worked answer
Let x and y be kilograms taken from X and Y before loss. Because the 2% loss is uniform, final mass is 0.98(x+y)=40, so x+y=40/0.98≈40.816. The nut fraction is unchanged because both nut mass and total mass are multiplied by 0.98: fraction=(0.30x+0.55y)/(x+y).
The 45% condition gives 0.30x+0.55y≥0.45(x+y), so 0.10y≥0.15x and y≥1.5x. The 48% condition gives 0.30x+0.55y≤0.48(x+y), so 0.07y≤0.18x and y≤(18/7)x≈2.571x. The stock condition y≤2x is stricter than the upper nut condition. Thus 1.5x≤y≤2x with x+y≈40.816.
From x+y≈40.816, the extremes give 40.816/3≤x≤40.816/2.5, or 13.61≤x≤16.33. A feasible choice is x=15.0 kg and y=25.816 kg. Then the nut fraction is (0.30·15+0.55·25.816)/40.816=(4.5+14.199)/40.816≈45.8%, which is inside 45–48%, and y=25.816≤2x=30.
2. The store can take mix only in 0.5 kg scoop increments and may discard excess after loss to leave exactly 40 kg. Find a whole-scoop feasible choice and verify it.
Read the worked answer
Choose x=16.0 kg and y=25.0 kg. The total before loss is 41.0 kg. After the uniform 2% loss, 0.98·41.0=40.18 kg remains, so discard 0.18 kg to leave exactly 40 kg.
Before discarding, the nut fraction is (0.30·16.0+0.55·25.0)/41.0=(4.8+13.75)/41.0=18.55/41.0≈45.24%, which is between 45% and 48%. The stock condition is 25.0≤2·16.0=32.0. Discarding a uniform portion does not change the percentage, so this whole-scoop blend is feasible.
3. Explain in one paragraph how the 2% loss affects the nut percentage and the amount that must be taken. What changes if the loss is not uniform?
Read the worked answer
A uniform loss cancels from the nut percentage because both the nut mass and the total mass are scaled by the same factor 0.98. It does affect the required input amount: x+y must be 40/0.98, not 40, so more mix must be taken than the final 40 kg. If the loss were not uniform, for example if bin Y lost 5% while bin X lost 2%, then the final fraction would be (0.98·0.30x+0.95·0.55y)/(0.98x+0.95y), and the earlier percentage constraints would have to be recalculated rather than using the original bin percentages directly.