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. Define variables: Let $x$ = kg of Food I, $y$ = kg of Food II. 2. **Constraints:** - Vitamin A: $2x + y \geq 8$ - Vitamin C: $x + 2y \geq 10$ - Non-negativity: $x \geq 0$, $y \geq 0$ 3. **Objective function:** Minimize cost $$Z = 50x + 70y$$ 4. **Graphical solution:** - From $2x + y = 8$, solve for $y$: $$y = 8 - 2x$$ - From $x + 2y = 10$, solve for $y$: $$y = \frac{10 - x}{2}$$ 5. **Find intersection point of constraints:** Multiply second equation by 2: $$2x + 4y = 20$$ Subtract first equation: $$(2x + 4y) - (2x + y) = 20 - 8 \Rightarrow 3y = 12 \Rightarrow y = 4$$ Substitute $y=4$ into $2x + y = 8$: $$2x + 4 = 8 \Rightarrow 2x = 4 \Rightarrow x = 2$$ 6. **Check corner points:** - $(0,5)$ from $x=0$ in $x + 2y = 10$ - $(4,0)$ from $y=0$ in $2x + y = 8$ - $(2,4)$ intersection point 7. **Evaluate cost at corners:** - $Z(0,5) = 50(0) + 70(5) = 350$ - $Z(4,0) = 50(4) + 70(0) = 200$ - $Z(2,4) = 50(2) + 70(4) = 100 + 280 = 380$ 8. **Feasibility check:** - $(0,5)$: Vitamin A $= 2(0) + 5 = 5 < 8$ (not feasible) - $(4,0)$: Vitamin C $= 4 + 0 = 4 < 10$ (not feasible) - $(2,4)$: satisfies both constraints (feasible) 9. **Check other points on constraints:** - On $2x + y = 8$, at $x=3$, $y=8 - 6 = 2$, Vitamin C $= 3 + 2(2) = 7 < 10$ (not feasible) - On $x + 2y = 10$, at $x=0$, $y=5$, Vitamin A $= 0 + 5 = 5 < 8$ (not feasible) 10. **Check points on axes:** - For Vitamin A: at $y=0$, $x \geq 4$; at $x=0$, $y \geq 8$ - For Vitamin C: at $y=0$, $x \geq 10$; at $x=0$, $y \geq 5$ 11. **Check other points:** - $(6,1)$: Vitamin A $= 2(6) + 1 = 13 \geq 8$ (ok), Vitamin C $= 6 + 2(1) = 8 < 10$ (no) - $(3,4)$: Vitamin A $= 2(3) + 4 = 10 \geq 8$ (ok), Vitamin C $= 3 + 2(4) = 11 \geq 10$ (ok), Cost $= 50(3) + 70(4) = 430$ - $(1,5)$: Vitamin A $= 2(1) + 5 = 7 < 8$ (no) - $(5,2)$: Vitamin A $= 2(5) + 2 = 12 \geq 8$ (ok), Vitamin C $= 5 + 2(2) = 9 < 10$ (no) - $(2,4)$ again feasible with cost 380 12. **Conclusion:** The minimum cost feasible solution is at $(2,4)$ with cost 380. **Final answer:** $$x=2, y=4, \quad \text{Minimum cost} = 380$$