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Question
Find the linear inequations for which the shaded area in Fig. 15.41 is the solution set. Draw the diagram of the solution set of the linear inequations:
Solution
Considering the line 2x + 3y = 6, we find that the shaded region and the origin (0, 0) are on the opposite side of this line and (0, 0) does not satisfy the inequation 2x + 3y\[\geq\]6 So, the first inequation is 2x + 3y\[\geq\]6
Considering the line 4x + 6y = 24, we find that the shaded region and the origin (0, 0) are on the same side of this line and (0, 0) satisfies the inequation 4x + 6y\[\leq\]24 So, the corresponding inequation is 4x + 6y \[\leq\] 24 Considering the line x \[-\]2y = 2, we find that the shaded region and the origin (0, 0) are on the same side of this line and (0, 0) satisfies the inequation x \[-\]2y\[\leq\] 2 So, the corresponding inequation is x\[-\] 2y\[\leq\]2Considering the line\[-\]3x + 2y = 3, we find that the shaded region and the origin (0, 0) are on the same side of this line and (0, 0) satisfies the inequation\[-\]3x + 2y\[\leq\]3 So, the corresponding inequation is\[-\]3x + 2y\[\leq\]3 Also, the shaded region is in the first quadrant. Therefore, we must have x \[\geq 0 \text{ and } y \geq 0\]Thus, the linear inequations comprising the given solution set are given below:
2x + 3y\[\geq\] 6, 4x + 6y\[\leq\] 24, x\[-\]2y\[\leq\]2,\[-\]3x + 2y\[\leq\] 3, x\[\geq 0 \text{ and } y \geq 0\]
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