1. **State the problem:** Jacob needs to buy meat and cheese to meet minimum weekly requirements of 12 units of carbohydrates and 8 units of protein. Meat contains 2 units carbs and 4 units protein per pound, cheese contains 3 units carbs and 1 unit protein per pound. Meat costs 3.5 per pound, cheese costs 2.2 per pound. We want to find how many pounds of meat and cheese minimize cost while meeting requirements. 2. **Define variables:** Let $x$ = pounds of meat, $y$ = pounds of cheese. 3. **Write constraints from requirements:** Carbohydrates: $2x + 3y \geq 12$ Protein: $4x + y \geq 8$ Also, $x \geq 0$, $y \geq 0$ since pounds can't be negative. 4. **Write cost function to minimize:** $$C = 3.5x + 2.2y$$ 5. **Solve constraints as equalities to find corner points:** From carbs: $2x + 3y = 12 \Rightarrow y = \frac{12 - 2x}{3}$ From protein: $4x + y = 8 \Rightarrow y = 8 - 4x$ Set equal to find intersection: $$\frac{12 - 2x}{3} = 8 - 4x$$ Multiply both sides by 3: $$12 - 2x = 24 - 12x$$ Bring terms together: $$-2x + 12x = 24 - 12$$ $$10x = 12 \Rightarrow x = 1.2$$ Substitute back for $y$: $$y = 8 - 4(1.2) = 8 - 4.8 = 3.2$$ 6. **Check other corner points where constraints meet axes:** - When $x=0$, carbs: $3y \geq 12 \Rightarrow y \geq 4$ - When $y=0$, protein: $4x \geq 8 \Rightarrow x \geq 2$ 7. **Evaluate cost at corner points:** - At $(x,y) = (0,4)$: $C = 3.5(0) + 2.2(4) = 8.8$ - At $(x,y) = (2,0)$: $C = 3.5(2) + 2.2(0) = 7.0$ - At intersection $(1.2,3.2)$: $C = 3.5(1.2) + 2.2(3.2) = 4.2 + 7.04 = 11.24$ 8. **Check feasibility:** - $(0,4)$ meets carbs but protein: $4(0)+4=4 < 8$ no - $(2,0)$ meets protein but carbs: $2(2)+3(0)=4 < 12$ no - Intersection $(1.2,3.2)$ meets both exactly. 9. **Since $(1.2,3.2)$ is the only point meeting both constraints, it is the feasible solution.** 10. **Answer:** Jacob should buy **1.2 pounds of meat** and **3.2 pounds of cheese** to minimize cost. The minimum cost is **11.24**.