The problem asks us to determine how much of a 40-hectare land should be planted with coffee and vegetables to maximize profit, given that coffee yields a profit of K2000 per hectare and requires 2 laborers per hectare, while vegetables yield a profit of K6000 per hectare and require 8 laborers per hectare. The total labor force is 120.

1. Problem Description

The problem asks us to determine how much of a 40-hectare land should be planted with coffee and vegetables to maximize profit, given that coffee yields a profit of K2000 per hectare and requires 2 laborers per hectare, while vegetables yield a profit of K6000 per hectare and require 8 laborers per hectare. The total labor force is

1

2

0.

2. Solution Steps

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

The total land constraint is:

$x + y \le 40$

The total labor constraint is:

$2x + 8y \le 120$

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

$P = 2000x + 6000y$

We also have the non-negativity constraints:

$x \ge 0$

$y \ge 0$

We can simplify the labor constraint by dividing by 2:

$x + 4y \le 60$

Now we have the following linear programming problem:

Maximize $P = 2000x + 6000y$

Subject to:

$x + y \le 40$

$x + 4y \le 60$

$x \ge 0$

$y \ge 0$

First, let's find the vertices of the feasible region.

1. Intersection of $x + y = 40$ and $x + 4y = 60$:

Subtract the first equation from the second:

$3y = 20$

$y = \frac{20}{3}$

Substitute $y = \frac{20}{3}$ into $x + y = 40$:

$x + \frac{20}{3} = 40$

$x = 40 - \frac{20}{3} = \frac{120 - 20}{3} = \frac{100}{3}$

So the intersection point is $(\frac{100}{3}, \frac{20}{3})$.

2. Intersection of $x + y = 40$ and $x = 0$:

If $x = 0$, then $y = 40$. The intersection point is $(0, 40)$.

3. Intersection of $x + 4y = 60$ and $y = 0$:

If $y = 0$, then $x = 60$. But this point violates the constraint $x + y \le 40$.

4. Intersection of $x + 4y = 60$ and $x = 0$:

If $x = 0$, then $4y = 60$, so $y = 15$. The intersection point is $(0, 15)$.

5. Intersection of $x + y = 40$ and $y = 0$:

If $y = 0$, then $x = 40$. The intersection point is $(40, 0)$.

Now we evaluate the objective function at the vertices of the feasible region:

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

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

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

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

Since $\frac{320000}{3}$ is the largest profit, the optimal solution is to plant $\frac{100}{3}$ hectares of coffee and $\frac{20}{3}$ hectares of vegetables.

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.