Contextual Teaching and Learning Essay

Linear Programming

By

Mandy Stein


Linear programming is used to maximize profit or minimize costs in a variety of industries.

There are several components of linear programming:

Constraints - These are represented in the form of inequalities.  They represent limitations on materials, resources, or time.

Feasible Region - The solution set of the system of constraints.

Objective Function - The function to be minimized or maximized.  This function represents profit or cost.

Corner-Point Principle - The maximum and minimum values of an objective function occur at one of the vertices of the feasible region.


Problem:

A farmer has 90 acres on which to plant millet and alfalfa.  Seed for one acre of millet costs $4 and seed for one acre of alfalfa costs $6.  Labor costs for millet are $20 per acre and labor costs for alfalfa are $10 per acre.  The expected income for an acre of millet is $110 and the expected income for an acre of alfalfa is $150.  The farmer wants to spend no more than $480 for seed and $1400 for labor.  What should he plant to maximize his profit?

The objective function represents the profit the farmer will make from planting m acres of millet and a acres of alfalfa.

R = 110m + 150a

What the farmer plants is limited by these constraints:

Translating these constraints into inequalities gives us the following system of inequalities:

m + a < 90
4m + 6a < 480
20m + 10a < 1400
0 < m
0 < a

Next, we will graph the system of constraints to obtain the feasible region and locate the vertices. Note: The last two inequalities were added because a negative number of acres cannot be planted.

 

 

 

Based on the Corner-Point Principle, the maximum value will be located at one of the vertices.  Thus, we evaluate the objective function at each vertex to determine which combination produces the maximum profit.

Vertex (m,a)

Objective Function

(0,80)

R = 110(0) + 150(80) = 12000

(30,60)

R = 110(30) + 150(60) = 12300

(50,40)

R = 110(50) + 150(40) = 11500

(70,0)

R = 110(70) + 150(0) = 7700

We can see from the table that income is maximized when 30 acres of millet and 60 acres of alfalfa are planted.  The maximum income is $12,300.


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