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 by Buffalo is $100x$. - Each Kestel carries 80 men, so total men carried by Kestel is $80y$. - Total men to transport: $100x + 80y \geq 1000$. - Each Buffalo carries 6 tonnes, so total equipment by Buffalo is $6x$. - Each Kestel carries 10 tonnes, so total equipment by Kestel is $10y$. - Total equipment to transport: $6x + 10y \geq 84$. 4. **Write the cost function to minimize:** - Cost per Buffalo: 2100 - Cost per Kestel: 1800 - Total cost: $C = 2100x + 1800y$ 5. **Formulate the system:** $$\begin{cases} 100x + 80y \geq 1000 \\ 6x + 10y \geq 84 \end{cases}$$ 6. **Solve the system for equality to find boundary points:** From men constraint: $$100x + 80y = 1000 \implies 5x + 4y = 50 \implies y = \frac{50 - 5x}{4}$$ From equipment constraint: $$6x + 10y = 84 \implies 3x + 5y = 42 \implies y = \frac{42 - 3x}{5}$$ 7. **Find intersection point by equating $y$ values:** $$\frac{50 - 5x}{4} = \frac{42 - 3x}{5}$$ Multiply both sides by 20: $$5(50 - 5x) = 4(42 - 3x)$$ $$250 - 25x = 168 - 12x$$ $$250 - 168 = 25x - 12x$$ $$82 = 13x \implies x = \frac{82}{13} = 6.31$$ 8. **Find $y$ at $x=6.31$:** $$y = \frac{50 - 5(6.31)}{4} = \frac{50 - 31.55}{4} = \frac{18.45}{4} = 4.61$$ 9. **Check integer values near intersection:** Try $x=7$, $y=4$: - Men: $100(7) + 80(4) = 700 + 320 = 1020 \geq 1000$ (OK) - Equipment: $6(7) + 10(4) = 42 + 40 = 82 < 84$ (Not OK) Try $x=7$, $y=5$: - Men: $700 + 400 = 1100 \geq 1000$ (OK) - Equipment: $42 + 50 = 92 \geq 84$ (OK) Cost: $2100(7) + 1800(5) = 14700 + 9000 = 23700$ Try $x=6$, $y=5$: - Men: $600 + 400 = 1000 \geq 1000$ (OK) - Equipment: $36 + 50 = 86 \geq 84$ (OK) Cost: $2100(6) + 1800(5) = 12600 + 9000 = 21600$ Try $x=5$, $y=6$: - Men: $500 + 480 = 980 < 1000$ (Not OK) Try $x=8$, $y=4$: - Men: $800 + 320 = 1120 \geq 1000$ (OK) - Equipment: $48 + 40 = 88 \geq 84$ (OK) Cost: $2100(8) + 1800(4) = 16800 + 7200 = 24000$ 10. **Minimum cost found at $x=6$, $y=5$ with cost 21600.** **Final answer:** Use 6 Buffalo aircraft and 5 Kestel aircraft to minimize cost while transporting all men and equipment.