Mine A gives 1, 3 and 5 tons of high, medium and low grade ore daily; mine B gives 2 tons of each. Needing 80, 160 and 200 tons at Shs. 200,000 per mine-day, the cheapest schedule is:

1. Let $x$ and $y$ be days worked by mines A and B. Minimise $f(x,y)=200{,}000(x+y)$. 2. Ore requirements give $x+2y\ge 80$, $3x+2y\ge 160$, $5x+2y\ge 200$, with $x,y\ge 0$. 3. Graphing the feasible region gives corner points $(80,0)$, $(40,20)$, $(20,50)$ and $(0,100)$. 4. Evaluating $x+y$ at each: $80$, $60$, $70$, $100$ mine-days. 5. The smallest is $60$ at $(40,20)$, a cost of $\text{Tsh }12{,}000{,}000$. 6. The other three options are the remaining corner points, each feasible but dearer. _Source: NECTA ACSEE 2023 Advanced Mathematics 142/1, Question 3: Linear Programming_