1. **Problem Statement:** A farmer wants to mix two types of food, A and B, to meet minimum daily nutritional requirements at minimum cost. 2. **Define variables:** Let $x$ = number of bags of Food A Let $y$ = number of bags of Food B 3. **Cost function to minimize:** $$\text{Cost} = 20000x + 22000y$$ 4. **Nutritional constraints:** - Proteins: $$40x + 30y \geq 150$$ - Minerals: $$20x + 20y \geq 90$$ - Vitamins: $$10x + 30y \geq 60$$ 5. **Solve constraints for $y$:** - From proteins: $$40x + 30y \geq 150 \Rightarrow 30y \geq 150 - 40x \Rightarrow y \geq \frac{150 - 40x}{30}$$ - From minerals: $$20x + 20y \geq 90 \Rightarrow 20y \geq 90 - 20x \Rightarrow y \geq \frac{90 - 20x}{20} = \frac{90 - 20x}{20}$$ - From vitamins: $$10x + 30y \geq 60 \Rightarrow 30y \geq 60 - 10x \Rightarrow y \geq \frac{60 - 10x}{30}$$ 6. **Find feasible $(x,y)$ pairs that satisfy all constraints and minimize cost:** Check intersection points of constraints: - Solve proteins and minerals equality: $$40x + 30y = 150$$ $$20x + 20y = 90$$ Multiply second by 1.5: $$30x + 30y = 135$$ Subtract from first: $$40x + 30y - (30x + 30y) = 150 - 135 \Rightarrow 10x = 15 \Rightarrow x = 1.5$$ Substitute back: $$20(1.5) + 20y = 90 \Rightarrow 30 + 20y = 90 \Rightarrow 20y = 60 \Rightarrow y = 3$$ - Check vitamins constraint at $(1.5,3)$: $$10(1.5) + 30(3) = 15 + 90 = 105 \geq 60$$ (satisfied) 7. **Calculate cost at $(1.5,3)$:** $$20000(1.5) + 22000(3) = 30000 + 66000 = 96000$$ 8. **Check other vertices for minimum cost:** - Intersection of proteins and vitamins: $$40x + 30y = 150$$ $$10x + 30y = 60$$ Subtract second from first: $$30x = 90 \Rightarrow x = 3$$ Substitute back: $$10(3) + 30y = 60 \Rightarrow 30 + 30y = 60 \Rightarrow 30y = 30 \Rightarrow y = 1$$ Cost: $$20000(3) + 22000(1) = 60000 + 22000 = 82000$$ - Intersection of minerals and vitamins: $$20x + 20y = 90$$ $$10x + 30y = 60$$ Multiply second by 2: $$20x + 60y = 120$$ Subtract first: $$20x + 60y - (20x + 20y) = 120 - 90 \Rightarrow 40y = 30 \Rightarrow y = 0.75$$ Substitute back: $$20x + 20(0.75) = 90 \Rightarrow 20x + 15 = 90 \Rightarrow 20x = 75 \Rightarrow x = 3.75$$ Cost: $$20000(3.75) + 22000(0.75) = 75000 + 16500 = 91500$$ 9. **Minimum cost is at $(3,1)$ with cost 82000.** 10. **Savings calculation:** Current purchase: 5 bags A and 4 bags B Current cost: $$20000(5) + 22000(4) = 100000 + 88000 = 188000$$ Recommended cost: $$82000$$ Savings: $$188000 - 82000 = 106000$$ 11. **Limitations of the technique:** - Assumes linear relationships and fixed nutrient content. - Ignores variability in feed quality and animal needs. - Does not consider storage, availability, or other practical constraints. 12. **Overcoming limitations:** - Use more complex models (non-linear programming). - Incorporate stochastic or fuzzy models to handle uncertainty. - Combine with expert knowledge and real-world testing. Final answers: - a) Buy 3 bags of Food A and 1 bag of Food B daily. - b) Savings of 106000 by switching. - c) Limitations and improvements as above.