suppose a farmer has100 acre farm and wants to maximize profit by allocating land to two type 0f crops, wheat and corn. the farmer has 200 units of labor and150 units of fertilizer available ,each acre of wheat requires 2units of and labor and 3 units of fertilizer while each acre of corn requires 4 units of labor and 2 units of fertilizer, the profit per acre for wheat is k300 and for corn is k400,find the optimal crop allocation
Let's assume the farmer allocates x acres of land to wheat and y acres of land to corn.
The objective is to maximize profit, which can be represented as:
Profit = 300x + 400y
However, there are constraints to consider:
1. Land constraint: x + y ≤ 100 (Total land available is 100 acres)
2. Labor constraint: 2x + 4y ≤ 200 (Total labor available is 200 units)
3. Fertilizer constraint: 3x + 2y ≤ 150 (Total fertilizer available is 150 units)
To find the optimal crop allocation, we need to solve this linear programming problem:
Maximize: 300x + 400y
Subject to:
x + y ≤ 100
2x + 4y ≤ 200
3x + 2y ≤ 150
To simplify the problem, we can divide all the constraints by their respective coefficients:
x + y ≤ 100
x + 2y ≤ 100
3x + y ≤ 75
Now we can plot these lines on a graph:
Line 1: x + y = 100 (blue line)
Line 2: x + 2y = 100 (green line)
Line 3: 3x + y = 75 (red line)
Finding the intersection points:
By solving the linear equations, the intersection points are:
A: (0, 75)
B: (25, 75)
C: (50, 50)
D: (75, 0)
The feasible region is the shaded region enclosed by these lines.
Next, we evaluate the objective function (profit) at each corner point of the feasible region:
A: 300(0) + 400(75) = 30,000
B: 300(25) + 400(75) = 45,000
C: 300(50) + 400(50) = 35,000
D: 300(75) + 400(0) = 22,500
From the evaluations, we see that the maximum profit of 45,000 (k45,000) is achieved when allocating 25 acres to wheat and 75 acres to corn.