1. **Problem Statement:** A cook wants to mix Food I and Food II to get at least 8 units of vitamin A and 10 units of vitamin C. 2. **Define variables:** Let $x$ = kg of Food I, $y$ = kg of Food II. 3. **Constraints from vitamin content:** - Vitamin A: $2x + y \geq 8$ - Vitamin C: $x + 2y \geq 10$ 4. **Cost function to minimize:** $$ \text{Cost} = 50x + 70y $$ 5. **Non-negativity constraints:** $$ x \geq 0, y \geq 0 $$ 6. **Graphical solution:** - Plot lines $2x + y = 8$ and $x + 2y = 10$. - Feasible region is where both inequalities hold. 7. **Find intersection points:** Solve system: $$ 2x + y = 8 $$ $$ x + 2y = 10 $$ 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 first: $$ 2x + 4 = 8 \Rightarrow 2x = 4 \Rightarrow x = 2 $$ 8. **Evaluate cost at vertices:** - At $(0,5)$ (from $x+2y=10$): Cost = $50(0)+70(5)=350$ - At $(4,0)$ (from $2x+y=8$): Cost = $50(4)+70(0)=200$ - At $(2,4)$ intersection: Cost = $50(2)+70(4)=100+280=380$ 9. **Check feasibility:** Points must satisfy both constraints. - $(0,5)$: $2(0)+5=5<8$ no - $(4,0)$: $4+2(0)=4<10$ no - $(2,4)$ satisfies both. 10. **Check corner points on axes:** - For $2x + y \geq 8$, if $y=0$, $x \geq 4$. - For $x + 2y \geq 10$, if $x=0$, $y \geq 5$. 11. **Feasible region vertices:** - $(4,3)$ satisfies $2(4)+3=11 \geq 8$ and $4+2(3)=10 \geq 10$ - $(6,1)$ satisfies $2(6)+1=13 \geq 8$ and $6+2(1)=8<10$ no 12. **Evaluate cost at $(4,3)$:** Cost = $50(4)+70(3)=200+210=410$ 13. **Minimum cost at intersection $(2,4)$ is 380 but check if feasible:** $2(2)+4=8$ yes, $2+2(4)=10$ yes, so feasible. 14. **Conclusion:** Minimum cost is Rs 380 at $x=2$ kg Food I and $y=4$ kg Food II.