1. **State the problem:** We need to transport 1000 men and 84 tonnes of equipment using two types of aircraft: Buffalo and Kestel. 2. **Define variables:** Let $x$ be the number of Buffalo aircraft and $y$ be the number of Kestel aircraft. 3. **Write constraints based on capacity:** - Each Buffalo carries 100 men, so total men carried: $100x$. - Each Kestel carries 80 men, so total men carried: $80y$. - Total men to move: 1000, so $$100x + 80y \geq 1000$$ - Each Buffalo carries 6 tonnes, so total equipment carried: $6x$. - Each Kestel carries 10 tonnes, so total equipment carried: $10y$. - Total equipment to move: 84 tonnes, so $$6x + 10y \geq 84$$ 4. **Cost function:** The cost per Buffalo is not given, but cost per Kestel is 1800. Since only Kestel cost is given, assume Buffalo cost is $c$ (unknown). To minimize total cost, $$\text{Cost} = c x + 1800 y$$ Since Buffalo cost is not given, we assume it is less or equal to Kestel or we minimize $y$ to reduce cost. 5. **Solve constraints for integer solutions:** From men constraint: $$100x + 80y \geq 1000$$ Divide by 20: $$5x + 4y \geq 50$$ From equipment constraint: $$6x + 10y \geq 84$$ Divide by 2: $$3x + 5y \geq 42$$ 6. **Find integer solutions $(x,y)$ satisfying both inequalities and minimizing $y$ (to minimize cost):** Try $y=6$: - Men: $5x + 4(6) = 5x + 24 \geq 50 \Rightarrow 5x \geq 26 \Rightarrow x \geq 6$. - Equipment: $3x + 5(6) = 3x + 30 \geq 42 \Rightarrow 3x \geq 12 \Rightarrow x \geq 4$. So for $y=6$, $x$ must be at least 6. Try $y=5$: - Men: $5x + 20 \geq 50 \Rightarrow 5x \geq 30 \Rightarrow x \geq 6$. - Equipment: $3x + 25 \geq 42 \Rightarrow 3x \geq 17 \Rightarrow x \geq 6$. Try $y=4$: - Men: $5x + 16 \geq 50 \Rightarrow 5x \geq 34 \Rightarrow x \geq 7$. - Equipment: $3x + 20 \geq 42 \Rightarrow 3x \geq 22 \Rightarrow x \geq 8$. Try $y=3$: - Men: $5x + 12 \geq 50 \Rightarrow 5x \geq 38 \Rightarrow x \geq 8$. - Equipment: $3x + 15 \geq 42 \Rightarrow 3x \geq 27 \Rightarrow x \geq 9$. Try $y=2$: - Men: $5x + 8 \geq 50 \Rightarrow 5x \geq 42 \Rightarrow x \geq 9$. - Equipment: $3x + 10 \geq 42 \Rightarrow 3x \geq 32 \Rightarrow x \geq 11$. Try $y=1$: - Men: $5x + 4 \geq 50 \Rightarrow 5x \geq 46 \Rightarrow x \geq 10$. - Equipment: $3x + 5 \geq 42 \Rightarrow 3x \geq 37 \Rightarrow x \geq 13$. Try $y=0$: - Men: $5x \geq 50 \Rightarrow x \geq 10$. - Equipment: $3x \geq 42 \Rightarrow x \geq 14$. 7. **Calculate total cost for each feasible pair $(x,y)$ assuming Buffalo cost is less than or equal to Kestel cost:** - For $y=5$, $x=6$: cost = $c \times 6 + 1800 \times 5 = 6c + 9000$ - For $y=6$, $x=6$: cost = $6c + 10800$ - For $y=4$, $x=8$: cost = $8c + 7200$ - For $y=3$, $x=9$: cost = $9c + 5400$ - For $y=2$, $x=11$: cost = $11c + 3600$ - For $y=1$, $x=13$: cost = $13c + 1800$ - For $y=0$, $x=14$: cost = $14c$ Since Buffalo cost $c$ is unknown, if $c < 1800$, minimizing $y$ reduces cost. 8. **Conclusion:** To minimize cost, use $x=13$ Buffalo and $y=1$ Kestel aircraft. **Final answer:** $$\boxed{x=13,\quad y=1}$$ This combination moves at least 1000 men and 84 tonnes equipment with minimum Kestel aircraft to reduce cost.