1. **State the problem:** We need to find (a) the total cost to pay for 1200 bags of produce based on past transport costs, and (b) the number of trips each vehicle (Pickup and Isuzu Diana) should make to transport more than 120 bags with minimum transport cost, given constraints. 2. **Part (a): Calculate cost for 1200 bags** - From January: 300 bags at 4500 per bag - From February: 700 bags at 8500 per bag - We assume the cost per bag is a weighted average based on these quantities. 3. **Calculate weighted average cost per bag:** $$\text{Average cost} = \frac{300 \times 4500 + 700 \times 8500}{300 + 700} = \frac{1350000 + 5950000}{1000} = \frac{7300000}{1000} = 7300$$ 4. **Calculate total cost for 1200 bags:** $$\text{Total cost} = 1200 \times 7300 = 8760000$$ 5. **Part (b): Minimize transport cost** - Let $x$ = number of Pickup trips, $y$ = number of Isuzu Diana trips - Pickup capacity = 18 bags/trip - Isuzu Diana capacity = 30 bags/trip - Total bags transported: $$18x + 30y > 120$$ - Cost per trip: Pickup = 240000, Isuzu Diana = 300000 - Total transport budget: 2400000 - Constraint on trips: $$x - y \leq 2$$ 6. **Formulate inequalities:** $$18x + 30y > 120$$ $$240000x + 300000y \leq 2400000$$ $$x - y \leq 2$$ 7. **Simplify transport budget inequality by dividing by 60000:** $$4x + 5y \leq 40$$ 8. **Find integer solutions $(x,y)$ satisfying:** - $18x + 30y > 120$ - $4x + 5y \leq 40$ - $x - y \leq 2$ 9. **Check feasible integer pairs:** Try $y=6$: - $x - 6 \leq 2 \Rightarrow x \leq 8$ - $4x + 5(6) \leq 40 \Rightarrow 4x + 30 \leq 40 \Rightarrow 4x \leq 10 \Rightarrow x \leq 2.5$ - So $x \leq 2.5$ and $x \leq 8$ means $x \leq 2$ - Check bags: $18(2) + 30(6) = 36 + 180 = 216 > 120$ (valid) - Total cost: $240000(2) + 300000(6) = 480000 + 1800000 = 2280000 \leq 2400000$ (valid) Try $x=2$, $y=6$ is feasible. 10. **Check if cost can be minimized further:** Try $x=1$, $y=7$: - $x - y = 1 - 7 = -6 \leq 2$ (valid) - $4(1) + 5(7) = 4 + 35 = 39 \leq 40$ (valid) - Bags: $18(1) + 30(7) = 18 + 210 = 228 > 120$ (valid) - Cost: $240000(1) + 300000(7) = 240000 + 2100000 = 2340000$ (higher than 2280000) Try $x=0$, $y=8$: - $x - y = -8 \leq 2$ (valid) - $4(0) + 5(8) = 40 \leq 40$ (valid) - Bags: $18(0) + 30(8) = 240 > 120$ (valid) - Cost: $0 + 2400000 = 2400000$ (higher than 2280000) 11. **Conclusion:** Minimum cost is at $x=2$, $y=6$ with total cost 2280000. **Final answers:** - (a) Total cost for 1200 bags = **8760000** - (b) Number of trips: Pickup = **2**, Isuzu Diana = **6** to minimize transport cost within budget.