1. **Problem Statement:** A farmer wants to minimize the daily cost of chicken feed using two types of feed, A and B. Feed A costs 20 per kg and contains 3000 units of vitamin V1 and 1000 units of vitamin V2 per kg. Feed B costs 40 per kg and contains 4000 units of vitamin V1 and 4000 units of vitamin V2 per kg. The chickens need at least 36000 units of vitamin V1 and 20000 units of vitamin V2 daily. We want to find how many kg of each feed to use to minimize cost while meeting vitamin requirements. 2. **Define variables:** Let $x$ = kg of Feed A Let $y$ = kg of Feed B 3. **Formulate constraints:** Vitamin V1 requirement: $$3000x + 4000y \geq 36000$$ Vitamin V2 requirement: $$1000x + 4000y \geq 20000$$ Also, $x \geq 0$, $y \geq 0$ since negative feed is impossible. 4. **Objective function (cost to minimize):** $$Z = 20x + 40y$$ 5. **Solve constraints for boundary lines:** From V1: $$3000x + 4000y = 36000 \Rightarrow 3x + 4y = 36$$ From V2: $$1000x + 4000y = 20000 \Rightarrow x + 4y = 20$$ 6. **Find intersection points of constraints:** Solve system: $$3x + 4y = 36$$ $$x + 4y = 20$$ Subtract second from first: $$2x = 16 \Rightarrow x = 8$$ Plug back: $$8 + 4y = 20 \Rightarrow 4y = 12 \Rightarrow y = 3$$ 7. **Check corner points (feasible region vertices):** - Point A: Intersection $(8,3)$ - Point B: Intersection with $x$-axis for V1: set $y=0$ in $3x+4y=36$ gives $x=12$ - Point C: Intersection with $x$-axis for V2: set $y=0$ in $x+4y=20$ gives $x=20$ - Point D: Intersection with $y$-axis for V1: set $x=0$ in $3x+4y=36$ gives $y=9$ - Point E: Intersection with $y$-axis for V2: set $x=0$ in $x+4y=20$ gives $y=5$ Check which points satisfy both constraints: - $(8,3)$ satisfies both - $(12,0)$: check V2: $1000*12 + 4000*0=12000<20000$ no - $(0,9)$: check V2: $1000*0 + 4000*9=36000>20000$ yes; check V1: $3000*0+4000*9=36000>36000$ yes - $(0,5)$: check V1: $3000*0+4000*5=20000<36000$ no Feasible corner points are $(8,3)$ and $(0,9)$. 8. **Calculate cost at feasible points:** At $(8,3)$: $$Z=20(8)+40(3)=160+120=280$$ At $(0,9)$: $$Z=20(0)+40(9)=360$$ 9. **Conclusion:** Minimum cost is 280 at $x=8$ kg of Feed A and $y=3$ kg of Feed B. **Final answer:** Use 8 kg of Feed A and 3 kg of Feed B daily to minimize cost. Minimum daily cost is 280.