1. **Problem Statement:** We have two machines C1 and C2 producing two grades of tyres A and B. Machine C1 produces 50 units of A and 30 units of B per hour, and machine C2 produces 30 units of A and 40 units of B per hour. The production requirements are at least 1400 units of A and 1200 units of B. The cost per hour for C1 is 50 and for C2 is 80. We want to minimize the total operating cost while meeting production requirements. 2. **Formulating the Linear Programming Problem (LPP):** Let $x$ = hours machine C1 operates Let $y$ = hours machine C2 operates Objective function (minimize cost): $$\text{Minimize } Z = 50x + 80y$$ Subject to production constraints: $$50x + 30y \geq 1400 \quad \text{(Grade A requirement)}$$ $$30x + 40y \geq 1200 \quad \text{(Grade B requirement)}$$ Non-negativity constraints: $$x \geq 0, \quad y \geq 0$$ 3. **Graphical Method Setup:** Rewrite inequalities as equalities to plot lines: $$50x + 30y = 1400$$ $$30x + 40y = 1200$$ Find intercepts: - For $50x + 30y = 1400$: - $x$-intercept: $x=\frac{1400}{50}=28$ - $y$-intercept: $y=\frac{1400}{30} \approx 46.67$ - For $30x + 40y = 1200$: - $x$-intercept: $x=\frac{1200}{30}=40$ - $y$-intercept: $y=\frac{1200}{40}=30$ 4. **Feasible Region:** The feasible region is where both inequalities are satisfied (above or on both lines) and $x,y \geq 0$. 5. **Find Corner Points:** - Intersection of constraints: Solve system: $$50x + 30y = 1400$$ $$30x + 40y = 1200$$ Multiply second equation by 5 and first by 3: $$150x + 120y = 6000$$ $$150x + 90y = 4200$$ Subtract: $$30y = 1800 \Rightarrow y = 60$$ Substitute $y=60$ into first equation: $$50x + 30(60) = 1400 \Rightarrow 50x + 1800 = 1400 \Rightarrow 50x = -400 \Rightarrow x = -8$$ Since $x$ cannot be negative, no feasible intersection here. Check intercept points: - At $x=0$, $y$ must satisfy both: - $30y \geq 1400 \Rightarrow y \geq 46.67$ - $40y \geq 1200 \Rightarrow y \geq 30$ So $y \geq 46.67$ - At $y=0$, $x$ must satisfy both: - $50x \geq 1400 \Rightarrow x \geq 28$ - $30x \geq 1200 \Rightarrow x \geq 40$ So $x \geq 40$ 6. **Evaluate Objective at Corner Points:** - Point A: $(0, 46.67)$ $$Z = 50(0) + 80(46.67) = 3733.6$$ - Point B: $(40, 0)$ $$Z = 50(40) + 80(0) = 2000$$ 7. **Check feasibility of intersection point:** Since intersection point has negative $x$, it is not feasible. 8. **Conclusion:** Minimum cost is at $(40,0)$ with cost $Z=2000$. **Final answer:** $$\boxed{\text{Operate machine C1 for } 40 \text{ hours and machine C2 for } 0 \text{ hours to minimize cost } Z=2000}$$