1, a farmer has 20 hectares for growing barley and wheat.the farmer has to decide how much of each to grow. The cost per hectare for barley is 30 and for wheat is 30. The farmer budgeted 480. Barley requires 1 man day per hectare and wheat requires 2 man days per hectare. there are 36 man-days available.the profit on barley is 100 per hectare and on wheat is 120 per hectare. Find the numbers of hectares of each the farmer should sow to maximize profits.
Let x be the number of hectares of barley to grow and y be the number of hectares of wheat to grow.
The objective is to maximize profit, which can be represented as a function:
Profit = 100x + 120y
Subject to the following constraints:
1) The total area of barley and wheat should not exceed 20 hectares:
x + y ≤ 20
2) The total cost should not exceed the budget of 480:
30x + 30y ≤ 480
3) The total man-days used should not exceed the available 36 man-days:
x + 2y ≤ 36
To solve this linear programming problem, we need to graph the feasible region and find the coordinates that maximize the objective function within that region.
First, let's find the vertices of the feasible region by solving the system of equations formed by the constraints:
1) x + y = 20 (convert the inequality to an equation)
2) 30x + 30y = 480
3) x + 2y = 36
Solving these equations, we get:
x = 8, y = 12 (intersection of x + y = 20 and 30x + 30y = 480)
x = 0, y = 20 (intersection of x + y = 20 and x + 2y = 36)
x = 20, y = 0 (intersection of x + y = 20 and x + 2y = 36)
Now we can calculate the profit (objective function) at each vertex:
Profit at (8, 12) = 100(8) + 120(12) = 2000
Profit at (0, 20) = 100(0) + 120(20) = 2400
Profit at (20, 0) = 100(20) + 120(0) = 2000
The maximum profit is at (0, 20) with 2400 profit.
Therefore, the farmer should sow 0 hectares of barley and 20 hectares of wheat to maximize profits.
To solve this problem, we can use linear programming to maximize profits.
Let's assign the following variables:
Let x = number of hectares of barley to sow
Let y = number of hectares of wheat to sow
We need to maximize the profit function:
Profit = 100x + 120y
Subject to the following constraints:
1. Total budget constraint: 30x + 30y = 480
2. Man-days constraint: x + 2y <= 36
3. Non-negativity constraint: x >= 0, y >= 0
Let's solve this problem step by step.
Step 1: Re-arrange the budget constraint equation:
30x + 30y = 480
Divide by 30:
x + y = 16 ----- (Equation 1)
Step 2: Set up the mathematical model:
Maximize: Profit = 100x + 120y
Subject to:
x + y = 16 ----- (Equation 1)
x + 2y <= 36
x >= 0, y >= 0
Step 3: Graph the feasible region:
To graph the feasible region, we need to find the intersection points of the lines described by the constraints.
Equation 1: x + y = 16
At x = 0, y = 16
At y = 0, x = 16
x + 2y = 36
At x = 0, y = 18
At y = 0, x = 36
Graphing these points and connecting the lines, we get a feasible region that is a triangle with vertices at (0, 16), (16, 0), and (0, 18).
Step 4: Identify the corner points of the feasible region:
We need to evaluate the profit function at each corner point to find the maximum profit.
Corner point A: (0, 16)
Profit at A = 100(0) + 120(16) = 1920
Corner point B: (16, 0)
Profit at B = 100(16) + 120(0) = 1600
Corner point C: (0, 18)
Profit at C = 100(0) + 120(18) = 2160
Step 5: Compare the profits and select the optimal solution:
The maximum profit is at corner point C, (0, 18). This means the farmer should sow 0 hectares of barley and 18 hectares of wheat to maximize profits.
Therefore, the farmer should sow 0 hectares of barley and 18 hectares of wheat to maximize profits.