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Question
Solve the following systems of inequations graphically:
12x + 12y ≤ 840, 3x + 6y ≤ 300, 8x + 4y ≤ 480, x ≥ 0, y ≥ 0
Solution
Converting the inequations to equations, we obtain:
12x + 12y = 840, 3x + 6y = 300, 8x + 4y = 480, x = 0, y = 0
12x + 12y = 840: This line meets the x-axis at (70, 0) and y-axis at (0, 70). Draw a thick line through these points.
Now, we see that the origin (0, 0) satisfies the inequation 12x + 12y\[\leq\]840
Therefore the region containing the origin is the solution of the inequality 12x + 12y\[\leq\]840
3x + 6y =300: This line meets the x-axis at (100, 0) and y-axis at (0, 50). Draw a thick line through these points.
Now, we see that the origin (0, 0) satisfies the inequation 3x + 6y\[\leq\]300
Therefore, the region containing the origin is the solution of the inequality 3x + 6\[\leq\]300
8x + 4y = 480: This line meets the x-axis at (60, 0) and y-axis at (0, 120). Draw a thick line through these points.
Now, we see that the origin (0, 0) satisfies the inequation 8x + 4y\[\leq\]480 Therefore, the region containing the origin is the solution of the inequality 8x + 4y\[\leq\]480
Also, x\[\geq 0, y \geq 0\]represens the first quadrant. So, the solution set must lie in the first quadrant.
Hence, the solution to the inequalities is the intersection of the above three solutions.
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