2.2.1 Solution of a Maximization ModeJ
Example 2.2-1
This example solves the Reddy Mikks model of Example 2.1-1.
Step 1. Determination ofthe Feasible Solution Space:
First, we account for the nonnegativity constraints Xl ~ 0 and X2 2: O. In Figure 2.1,
the horizontal axis Xl and the vertical axis X2 represent the exterior- and interior-paint
variables, respectively. Thus, the nonnegativity of the variables restricts the solutionspace
area to the first quadrant that lies above the xl-axis and to the right of the
x2-axis.
To account for the remaining four constraints, first replace each inequality
with an equation and then graph the resulting straight line by locating two distinct
points on it. For example, after replacing 6x[ + 4X2 :::; 24 with the straight line
6xl + 4x2 = 24, we can determine two distinct points by first setting XI = 0 to
obtain X2 = ¥= 6 and then setting X2 = 0 to obtain XI = ~ = 4. Thus, the line
passes through the two points (0,6) and (4,0), as shown by line (1) in Figure 2.1.
Next, consider the effect of the inequality. All it does is divide the (xJ, x2)-plane
into two half-spaces, one on each side of the graphed line. Only one of these two
halves satisfies the inequality. To determine the correct side, choose (0,0) as a
reference point. If it satisfies the inequality, then the side in which it lies is the
FIGURE 2.1
Feasible space of the Reddy Mikks model
2.2.1 Solution of a Maximization ModeJExample 2.2-1This example solves the Reddy Mikks model of Example 2.1-1.Step 1. Determination ofthe Feasible Solution Space:First, we account for the nonnegativity constraints Xl ~ 0 and X2 2: O. In Figure 2.1,the horizontal axis Xl and the vertical axis X2 represent the exterior- and interior-paintvariables, respectively. Thus, the nonnegativity of the variables restricts the solutionspacearea to the first quadrant that lies above the xl-axis and to the right of thex2-axis.To account for the remaining four constraints, first replace each inequalitywith an equation and then graph the resulting straight line by locating two distinctpoints on it. For example, after replacing 6x[ + 4X2 :::; 24 with the straight line6xl + 4x2 = 24, we can determine two distinct points by first setting XI = 0 toobtain X2 = ¥= 6 and then setting X2 = 0 to obtain XI = ~ = 4. Thus, the linepasses through the two points (0,6) and (4,0), as shown by line (1) in Figure 2.1.Next, consider the effect of the inequality. All it does is divide the (xJ, x2)-planeinto two half-spaces, one on each side of the graphed line. Only one of these twohalves satisfies the inequality. To determine the correct side, choose (0,0) as areference point. If it satisfies the inequality, then the side in which it lies is theFIGURE 2.1Feasible space of the Reddy Mikks model
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