The villagers have 40 hectares of land. Planting coffee yields a profit of K2000 per hectare, and requires 2 workers per hectare. Planting vegetables yields a profit of K6000 per hectare, and requires 8 workers per hectare. The total labor force is 120. We need to determine the optimal allocation of land between coffee and vegetables to maximize profit.

1. Problem Description

The villagers have 40 hectares of land. Planting coffee yields a profit of K2000 per hectare, and requires 2 workers per hectare. Planting vegetables yields a profit of K6000 per hectare, and requires 8 workers per hectare. The total labor force is

1

2

0. We need to determine the optimal allocation of land between coffee and vegetables to maximize profit.

2. Solution Steps

Let $x$ be the number of hectares planted with coffee, and $y$ be the number of hectares planted with vegetables.

The objective is to maximize the profit $P$, which is given by:

$P = 2000x + 6000y$

We have the following constraints:

1. Land constraint: $x + y \le 40$

2. Labor constraint: $2x + 8y \le 120$

3. Non-negativity constraints: $x \ge 0$, $y \ge 0$

We simplify the labor constraint:

$x + 4y \le 60$

Now we solve graphically or by considering the vertices of the feasible region.

The vertices of the feasible region are the intersections of the constraint lines:

1. $x = 0$, $y = 0$: $(0, 0)$

2. $x = 0$, $x + 4y = 60$: $(0, 15)$

3. $y = 0$, $x + y = 40$: $(40, 0)$

4. $x + y = 40$, $x + 4y = 60$: Subtracting the first equation from the second, we get $3y = 20$, so $y = \frac{20}{3}$. Then $x = 40 - \frac{20}{3} = \frac{120 - 20}{3} = \frac{100}{3}$. So the intersection is $(\frac{100}{3}, \frac{20}{3})$

5. $x+4y = 60$, $x+y = 40$: $x = 60-4y$ then $60-4y+y = 40$, $20 = 3y$, $y=\frac{20}{3}$, $x = 40-\frac{20}{3} = \frac{100}{3}$ which is $(\frac{100}{3}, \frac{20}{3})$

Now we evaluate the profit function at these vertices:

1. $(0, 0)$: $P = 2000(0) + 6000(0) = 0$

2. $(0, 15)$: $P = 2000(0) + 6000(15) = 90000$

3. $(40, 0)$: $P = 2000(40) + 6000(0) = 80000$

4. $(\frac{100}{3}, \frac{20}{3})$: $P = 2000(\frac{100}{3}) + 6000(\frac{20}{3}) = \frac{200000 + 120000}{3} = \frac{320000}{3} \approx 106666.67$

Therefore, the maximum profit occurs at $x = \frac{100}{3}$ and $y = \frac{20}{3}$.

3. Final Answer

To maximize profit, the villagers should plant $\frac{100}{3}$ hectares of coffee and $\frac{20}{3}$ hectares of vegetables. This is approximately 33.33 hectares of coffee and 6.67 hectares of vegetables.