Trail-Mix Blends and a Missing Nut Delivery
A food cooperative packs three trail-mix blends, X, Y and Z, from oats, honey and nuts. Each batch of X uses 2 kg oats, 1 kg honey and 1 kg nuts; each batch of Y uses 1, 2, 1; each batch of Z uses 1, 1, 2. This week the warehouse has 104 kg oats, 96 kg honey and 88 kg nuts. The cooperative wants to use all three stocks exactly. A buyer asks whether a fourth blend W, using 1 kg of each ingredient, could be added and still use all stocks exactly. Later, a nut supplier fails to deliver, so no Z can be made; the cooperative still wants to use as much of the 88 kg of nuts as possible while not exceeding the oats and honey. The manager needs to know whether the production plan is unique, and what to do when the mix changes.
1. Write the system of equations, solve it using a matrix method, and state how many batches of each blend should be made. Verify the determinant argument for uniqueness.
Read the worked answer
Let x, y and z be batches of X, Y and Z. The stock equations are 2x + y + z = 104, x + 2y + z = 96, x + y + 2z = 88.
The coefficient matrix is A = [[2,1,1],[1,2,1],[1,1,2]]. Its determinant is 2(4−1) − 1(2−1) + 1(1−2) = 6 − 1 − 1 = 4 ≠ 0, so A is invertible and the solution is unique.
Subtracting equation 2 from equation 1 gives x − y = 8. Subtracting equation 3 from equation 2 gives y − z = 8. Hence x = y + 8 and z = y − 8. Substituting into equation 1: 2(y + 8) + y + (y − 8) = 104, so 4y + 8 = 104, y = 24. Then x = 32 and z = 16.
The plan is 32 batches of X, 24 of Y and 16 of Z. Check: oats 64 + 24 + 16 = 104; honey 32 + 48 + 16 = 96; nuts 32 + 24 + 32 = 88. The nonzero determinant guarantees that no other plan uses all three stocks exactly.
2. If blend W (1,1,1) is added, can all stocks be used exactly? How many solutions are there? Explain what this means for the manager.
Read the worked answer
Let w be batches of W. The system becomes 2x + y + z + w = 104, x + 2y + z + w = 96, x + y + 2z + w = 88.
There are 3 equations and 4 unknowns. The first three columns still form the matrix A with det 4 ≠ 0, so the coefficient matrix has rank 3 and the system is consistent for the given right-hand side. Therefore there are infinitely many solutions, not a unique plan.
Subtracting as before gives x − y = 8 and y − z = 8, so x = y + 8 and z = y − 8. Substituting into the first equation: 2(y + 8) + y + (y − 8) + w = 104, so 4y + 8 + w = 104 and w = 96 − 4y.
Nonnegativity requires y ≥ 8 (for z ≥ 0) and y ≤ 24 (for w ≥ 0), giving 17 integer-valued plans. For example, y = 16 gives x = 24, z = 8, w = 32; y = 20 gives x = 28, z = 12, w = 16. The manager cannot choose on stock balance alone and must use a secondary criterion such as profit, shelf life or demand.
3. With no Z available, set up the reduced system for X and Y only. Show whether all 88 kg of nuts can be used exactly and find the plan that uses all oats and honey and the maximum possible nuts.
Read the worked answer
Let x and y be batches of X and Y. The constraints are oats 2x + y ≤ 104, honey x + 2y ≤ 96, nuts x + y ≤ 88.
First test whether all nuts can be used: set x + y = 88. Then oats = 2x + y = x + 88 ≤ 104, so x ≤ 16. Honey = x + 2y = x + 2(88 − x) = 176 − x ≤ 96, so x ≥ 80. Both cannot hold at once. Therefore using all 88 kg of nuts exactly is impossible without exceeding honey or oats.
So maximise n = x + y subject to the two supply constraints. The vertices are (0,0), (52,0), (0,48) and the intersection of 2x + y = 104 with x + 2y = 96. Solving gives x = 112/3 ≈ 37.33 and y = 88/3 ≈ 29.33, with x + y = 200/3 ≈ 66.67. This is the maximum nut use.
At that point oats = 104 and honey = 96 are both fully used, and nuts used = 66.67 kg, leaving 88 − 66.67 = 21.33 kg nuts unused. The cooperative should make about 37.33 batches of X and 29.33 of Y, or scale to whole batches while accepting a small deviation.