Linear
Programming: Introduction
(page
1 of 5)

Sections: Optimizing linear systems,
Setting up word problems

Linear programming is the process of taking
various linear inequalities relating to some situation, and finding the
“best” value obtainable under those conditions. A typical example
would be taking the limitations of materials and labor, and then determining
the “best” production levels for maximal profits under those
conditions.

In “real life”, linear programming
is part of a very important area of mathematics called “optimization
techniques”. This field of study (or at least the applied results
of it) are used every day in the organization and allocation of resources.
These “real life” systems can have dozens or hundreds of variables,
or more. In algebra, though, you’ll only work with the simple (and graphable)
two-variable linear case.

The general process for solving linear-programming
exercises is to graph the inequalities (called the “constraints”)
to form a walled-off area on the
x,y-plane
(called the “feasibility region”). Then you figure out the coordinates
of the corners of this feasibility region (that is, you find the intersection
points of the various pairs of lines), and test these corner points in
the formula (called the “optimization equation”) for which you’re
trying to find the highest or lowest value.

• Find the maximal and
minimal value of
z
= 3
x + 4y

subject to the following constraints:
• The three inequalities in the curly braces
are the constraints. The area of the plane that they mark off will be
the feasibility region. The formula “
z
= 3x + 4y
” is
the optimization equation. I need to find the
(x,
y)
corner points of the feasibility
region that return the largest and smallest values of
z.

My first step is to solve each inequality
for the more-easily graphed equivalent forms: It’s easy to graph
the system
: To find the corner points — which aren’t
always clear from the graph — I’ll pair the lines (thus forming a system
of linear equations
) and solve:

 y y y = ( 1/2 )x = ( 1/2 )x = 3xy + 7y + 7y = x = 3x = x  2  2 ( 1/2 ( 1/2 3x )x )x = x + 7 = 3xx + 7 = x  22x + 14 = 6x14  2x = 2x = 7x2 + 14 = 2x = 1 = x  418 = 3x6 y y = x = 3(1) = 3 = 3(2) = 6 y = (6)  2 = 4 corner point at corner point at (6, corner pt. at (1, (2, 6) 4) 3)

So the corner points are (2,
6)
, (6,
4)
, and (1,
3)
.

Somebody really smart proved that, for
linear systems like this, the maximum and minimum values of the optimization
equation will always be on the corners of the feasibility region. So,
to find the solution to this exercise, I only need to plug these three
points into “
z
= 3x + 4y
“.

(2, 6):      z
= 3(2)   + 4(6)   =   6 + 24 =   30

(6, 4):
z
= 3(6)   + 4(4)   = 18 + 16 =   34

(1, 3):  z
= 3(1) + 4(3) = 3  12 = 15

Then the
maximum of
z
= 34
occurs
at
(6,
4)
,
and
the
minimum of
z
= 15
occurs
at
(1,
3)
.

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