1. **State the problem:** A farmer wants to buy bags of two types of animal feed, Type A and Type B, to meet minimum nutritional requirements and minimize cost. 2. **Define variables:** Let $x$ = number of Type A bags, $y$ = number of Type B bags. 3. **Write constraints based on oats and corn content:** - Oats: Type A has 3 pounds per bag, Type B has 1 pound per bag, and the mixture must have at least 18 pounds. $$3x + y \geq 18$$ - Corn: Type A has 3 pounds per bag, Type B has 7 pounds per bag, and the mixture must have at least 54 pounds. $$3x + 7y \geq 54$$ 4. **Stock constraints:** - Maximum 11 bags of Type A available: $$x \leq 11$$ - Maximum 12 bags of Type B available: $$y \leq 12$$ 5. **Non-negativity constraints:** $$x \geq 0, \quad y \geq 0$$ 6. **Cost function to minimize:** $$C = 3x + 5y$$ 7. **Solve the system:** - From $3x + y \geq 18$, express $y \geq 18 - 3x$. - From $3x + 7y \geq 54$, express $y \geq \frac{54 - 3x}{7}$. 8. **Check feasible integer points within stock limits:** - Try $x=6$: - $y \geq 18 - 3(6) = 0$ - $y \geq \frac{54 - 3(6)}{7} = \frac{54 - 18}{7} = \frac{36}{7} \approx 5.14$ So $y \geq 5.14$, minimum integer $y=6$. - Check cost at $(6,6)$: $$C = 3(6) + 5(6) = 18 + 30 = 48$$ - Try $x=7$: - $y \geq 18 - 21 = -3$ (so $y \geq 0$) - $y \geq \frac{54 - 21}{7} = \frac{33}{7} \approx 4.71$ So $y \geq 5$. - Cost at $(7,5)$: $$C = 3(7) + 5(5) = 21 + 25 = 46$$ - Try $x=8$: - $y \geq 18 - 24 = -6$ (so $y \geq 0$) - $y \geq \frac{54 - 24}{7} = \frac{30}{7} \approx 4.29$ So $y \geq 5$. - Cost at $(8,5)$: $$C = 3(8) + 5(5) = 24 + 25 = 49$$ - Try $x=5$: - $y \geq 18 - 15 = 3$ - $y \geq \frac{54 - 15}{7} = \frac{39}{7} \approx 5.57$ So $y \geq 6$. - Cost at $(5,6)$: $$C = 3(5) + 5(6) = 15 + 30 = 45$$ - Try $x=4$: - $y \geq 18 - 12 = 6$ - $y \geq \frac{54 - 12}{7} = \frac{42}{7} = 6$ So $y \geq 6$. - Cost at $(4,6)$: $$C = 3(4) + 5(6) = 12 + 30 = 42$$ - Try $x=3$: - $y \geq 18 - 9 = 9$ - $y \geq \frac{54 - 9}{7} = \frac{45}{7} \approx 6.43$ So $y \geq 9$. - Cost at $(3,9)$: $$C = 3(3) + 5(9) = 9 + 45 = 54$$ 9. **Check stock limits:** All these points satisfy $x \leq 11$ and $y \leq 12$. 10. **Find minimum cost:** Among tested points, minimum cost is at $(4,6)$ with cost 42. **Final answer:** The farmer should buy **4 bags of Type A** and **6 bags of Type B** to minimize cost while meeting nutritional and stock constraints.