College Algebra 9th txtbk

22 CHAPTER 7 SYSTEMS OF EQUATIONS AND MATRICES
We now have a mathematical model for the problem under con-
MATHEMATICAL MODEL
sideration:
Maximize
Subject to
P 40x 50y
5x 4y 200
2x 3y 108
x, y 0
Objective function
Problem constraints
Nonnegative constraints
GRAPHIC SOLUTION Solving the system of linear inequality constraints graphically, as in
Section 10-7, we obtain the feasible region for production schedules, as shown in Figure 1.
y
(0, 36)
Fabricating capacity line
5x 4y 200
All lines are restricted to the
first quadrant because of the
nonnegative constraints x, y 0.
20
Feasible
region
(24, 20)
Finishing capacity line
2x 3y 108
(0, 0)
20
(40, 0)
x
Z Figure 1
Any production schedule (x, y) in the feasible region is possible—but which is the best?
That is, which produces the largest profit? We could try some schedules in the region to
see what the profit is; for x 24 and y 10, the weekly profit is
P 40(24) 50(10) $1,460
For a different point in the feasible region, (15, 20), the profit is
P 40(15) 50(20) $1,600
But how do we know when we’ve found the largest profit? Such a schedule, if it exists, is
called an optimal solution to the problem because it produces the maximum value of the
objective function and is in the feasible region. It is not practical to use point-by-point
checking to find the optimal solution. Even if we consider only points with integer coordinates,
there are over 800 such points in the feasible region for this problem. Instead, we use
the theory that has been developed to solve linear programming problems. Using advanced
techniques, it can be shown that
Corner
Point
(x, y)
(0, 0) 0
(0, 36) 1,800
(24, 20) 1,960
(40, 0) 1,600
Objective
Function
P 40x 50y
Maximum
value of P
If the feasible region is bounded, then one or more of the corner points of the
feasible region is an optimal solution to the problem.
The maximum value of the objective function is unique; however, there can be more than
one feasible production schedule that will produce this unique value. We will have more to
say about this later in this section.
Since the feasible region for this problem is bounded, at least one of the corner points,
(0, 0), (0, 36), (24, 20), or (40, 0), is an optimal solution. To find which one, we evaluate
P 40x 50y at each corner point and choose the corner point that produces the
largest value of P. It is convenient to organize these calculations in a table, as shown in
the margin.