Questions 10 - 15 are based on the following information: A farmer in Egypt owns
ID: 3363300 • Letter: Q
Question
Questions 10 - 15 are based on the following information:
A farmer in Egypt owns 50 acres of land. He is going to plant each acre with cotton or corn. Each acre planted with cotton yields $400 profit; each with corn yields $200 profit. The labor and fertilizer used for each acre are given in the table below. Resources available include 150 workers and 200 tons of fertilizer. Cotton Corn Labor (Workers) 5 3 Fertilizer (Tons) 6 2 Formulate a linear programming model that will enable the farmer to determine the number of acres that should be planted cotton and/or corn in order to maximize his profit. Provide the objective function and constraints in the box below.
X1 - number of acres planted cotton
X2 - number of acres planted corn
Hint: You will need three constraints.
What are the optimal decision variable values?
What is the optimal objective function value?
Which of the following constraints is binding?
How much should the farmer be willing to pay for an additional worker?
How much should the farmer be willing to pay for an additional ton of fertilizer?
Explanation / Answer
The linear program consists of decision variables
Let x = acres of cotton to plant
Let y = acres of corn plant
An objective function - to maximize 400x + 200y
And constraints
x + y <= 50 (acres available)
5x + 3y <= 150 (labor)
6x + 2y <= 200 (fertilizer)
This can easily be solved using a solver such as LINDO using
max 400x + 200y
st
x + y <= 50
6x + 2y <= 200
5x + 3y <= 150
Since this problem contains only 2 variables it can also be easily solved
graphically as follows.
Graph all the constrains - this will create a convex feasible region. The optimal solution will be at
an extreme point of this region. Evaluate the objective function at each extreme point to determine the
maximum value.
Let's give it a try. To graph each line we need to find 2 points on each line. That can be done simply by
setting x to 0 and solving for y and vice-versa.
For x + y = 50, if x = 0, y = 50 and vice-
versa so we have the line defined by the points (0,50) and (50,0)
Now for 6x + 2y = 500 simplifying
2y = -6x + 200
y = -3x + 100 so if x = 0, y = 100 and if y = 0, x = 33.33
so that line is defined by the points (0,100) (33.33,0)
Next for 5x + 3y = 150 simplifying gives
3y = -5x + 150
y = -5x/3 + 50 so if x = 0, y = 50 and if y = 0, x = 30
so that line is defined by the points (0,50) anf (30,0)
Finally the point where x + y = 50 and 6x + 2y = 200 intersect can be found by
multiplying both sides of x + y = 50 by -2 and adding to 6x + 2y = 200
-2x -2y = -100
6x + 2y = 200 giving
4x = 100, or x = 25 and by substitution y also = 25 so we no have the point (25,25)
We are also bound by the x and y axis since we are assuming x and y must both be >= 0
so the origin (0,0) is also one of our points. Now lets evaluate at each point.
Note that we are only interested in the extreme point of the constrainted region so (0,100),(33.33,0) and (50,0)
are not in our feasible region, only the following 3 points.
(0,0) - 400 * 0 + 200 * 0 = 0
(0,50) - 400 * 0 + 200 * 50 = 10,000
(30,0) = 400 * 30 + 200 * 0 = 12,000
Solving yields an optimal solution of x =30, y = 0 for an objective value of 12,000
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