Câu hỏi:

Let $x$ be the number of type A cars and $y$ be the number of type B cars. The objective is to minimize the cost $C = 4x + 3y$ (in millions of VND), subject to the following constraints: * $20x + 10y \ge 100$ (number of people) * $0.5x + 2y \ge 6$ (tons of goods) * $0 \le x \le 8$ * $0 \le y \le 6$ * $x, y$ are integers The constraints can be simplified to: * $2x + y \ge 10$ * $x + 4y \ge 12$ * $0 \le x \le 8$ * $0 \le y \le 6$ We evaluate the cost at the corner points of the feasible region: 1. Intersection of $2x + y = 10$ and $x + 4y = 12$: Solving this system yields $x = 4$ and $y = 2$. Cost: $C = 4(4) + 3(2) = 22$ 2. Intersection of $2x + y = 10$ and $y = 0$: This gives $x = 5$ and $y = 0$. Cost: $C = 4(5) + 3(0) = 20$ 3. Consider the point $x=1, y=5$. $C = 4(1) + 3(5) = 19$, $2x+y = 7 < 10$. Invalid. 4. Consider the point $x=2, y=4$. $C = 4(2) + 3(4) = 20$, $2x+y = 8 < 10$. Invalid. 5. Consider the point $x=0, y=6$. C = $4(0) + 3(6) = 18$. $2x+y = 6 < 10$. Invalid. Let's test the given options: 1. $x=5, y=0$. $2(5) + 0 = 10 \ge 10$, $5 + 4(0) = 5 < 12$. $C = 4(5) = 20$. Invalid. 2. $x=0, y=6$. $0 + 6 = 6 \ge 10$ False, $0 + 4(6) = 24 \ge 12$. $C = 3(6) = 18$. Invalid. 3. $x=2, y=4$. $2(2) + 4 = 8 < 10$. $2 + 4(4) = 18 \ge 12$. $C = 4(2) + 3(4) = 20$. Invalid. 4. $x=1, y=5$. $2(1) + 5 = 7 < 10$. $1 + 4(5) = 21 \ge 12$. $C = 4(1) + 3(5) = 19$. Invalid. None of the provided options satisfy the constraints. There must be an error in the provided options or the problem setup. The correct option should have a cost around 20 million and should satisfy the above conditions.