1. **Problem Statement:** A cook wants to mix two foods to get at least 8 units of vitamin A and 10 units of vitamin C. Food I has 2 units/kg vitamin A and 1 unit/kg vitamin C. Food II has 1 unit/kg vitamin A and 2 units/kg vitamin C. Cost: Food I = 50 per kg, Food II = 70 per kg. 2. **Define variables:** Let $x$ = kg of Food I, $y$ = kg of Food II. 3. **Constraints:** Vitamin A: $2x + y \geq 8$ Vitamin C: $x + 2y \geq 10$ Non-negativity: $x \geq 0$, $y \geq 0$ 4. **Objective function:** Minimize cost $Z = 50x + 70y$ 5. **Graphical solution:** Plot constraints: - $2x + y = 8$ gives $y = 8 - 2x$ - $x + 2y = 10$ gives $y = \frac{10 - x}{2}$ 6. **Find intersection points:** - Intersection of $2x + y = 8$ and $x + 2y = 10$: Multiply second by 2: $2x + 4y = 20$ Subtract first: $(2x + 4y) - (2x + y) = 20 - 8 \Rightarrow 3y = 12 \Rightarrow y = 4$ Substitute $y=4$ into $2x + y = 8$: $2x + 4 = 8 \Rightarrow 2x = 4 \Rightarrow x = 2$ 7. **Check corner points:** - At $(0,5)$ from $x=0$ in $x + 2y = 10$ - At $(4,0)$ from $y=0$ in $2x + y = 8$ - At $(2,4)$ intersection point 8. **Evaluate cost at corners:** - $Z(0,5) = 50(0) + 70(5) = 350$ - $Z(4,0) = 50(4) + 70(0) = 200$ - $Z(2,4) = 50(2) + 70(4) = 100 + 280 = 380$ 9. **Feasibility check:** - $(0,5)$ satisfies $2(0)+5=5<8$ vitamin A constraint, so not feasible. - $(4,0)$ satisfies $2(4)+0=8$ vitamin A and $4+0=4<10$ vitamin C, not feasible. - $(2,4)$ satisfies both constraints. 10. **Check other feasible points on constraints:** - On $2x + y = 8$, check $x=3$, $y=8-6=2$: Vitamin C: $3 + 2(2) = 3 + 4 = 7 < 10$ no. - On $x + 2y = 10$, check $x=0$, $y=5$ vitamin A fails. 11. **Try intersection with axes:** - For vitamin A: $2x + y \\geq 8$, at $y=0$, $x \\geq 4$; at $x=0$, $y \\geq 8$. - For vitamin C: $x + 2y \\geq 10$, at $y=0$, $x \\geq 10$; at $x=0$, $y \\geq 5$. 12. **Check point $(6,1)$:** Vitamin A: $2(6)+1=13 \\geq 8$ ok Vitamin C: $6 + 2(1)=8 < 10$ no 13. **Check point $(3,4)$:** Vitamin A: $2(3)+4=10 \\geq 8$ ok Vitamin C: $3 + 2(4)=11 \\geq 10$ ok Cost: $50(3)+70(4)=150+280=430$ 14. **Check point $(1,5)$:** Vitamin A: $2(1)+5=7 < 8$ no 15. **Check point $(5,2)$:** Vitamin A: $2(5)+2=12 \\geq 8$ ok Vitamin C: $5 + 2(2)=9 < 10$ no 16. **Check point $(2,4)$ again:** Feasible and cost $380$. 17. **Conclusion:** The minimum cost feasible solution is at $(2,4)$ with cost $380$. **Final answer:** $$x=2, y=4, \quad \text{Minimum cost} = 380$$