Exercises

A: (x,y,z) = (3.2,6,0.6)

B: (x,y,z) = (0.4, 0.6, 1.2)

C: (x,y,z) = (6,1,8)

D: (x,y,z) = (100,200,500)

Investigation

Notice that changing the weight (20kg) in the first equation will not change the nature of the solutions. The new plane might have equation x + y + z = 25, but this is parallel to the original plane, so there should still be one solution.

However, trying x + y + z =25 yields the solution (x,y,z) = (-6.25, 43.75, -12.5). This requires buying a negative amount of Colombian and Kenyan coffee. So although there is a mathematical solution, it is not a solution to the problem.

Similarly, buying 15 kg of coffee yields the solution (x,y,z) = (16.25, -33.75, 32.5). Adam cannot make this purchase either.

Some exploration with possible numbers yields possible solutions from 19.5 kg of coffee (6.125, 1.125, 12.25) up to 22 kg of coffee (0.5, 20.5, 1).

If we make the weight of coffee purchased W kg, then the algebraic solution is:

x = 50 - 2.25W

y = 7.75W - 150

z = 100 - 4.5W

All of these must be non-negative (≥ 0).

50 ≥ 2.25W

7.75W ≥ 150

100 ≥ 4.5W

The first and third constraints are the same (W ≤ 22.22 kg), while the second is W ≥ 19.35 kg.

The range of possible weights to buy, with the other two equations met, is between 19.35 and 22.22 kg (2 d.p.).