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Question
Minimize z = 2x + 4y is subjected to 2x + y ≥ 3, x + 2y ≥ 6, x ≥ 0, y ≥ 0 show that the minimum value of z occurs at more than two points
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Solution
To draw the feasible region, construct table as follows:
| Inequality | 2x + y ≥ 3 | x + 2y ≥ 6 |
| Corresponding equation (of line) | 2x + y = 3 | x + 2y = 6 |
| Intersection of line with X-axis | `(3/2, 0)` | (6, 0) |
| Intersection of line with Y-axis | (0, 3) | (0, 3) |
| Region | Non-origin side | Non-origin side |
x ≥ 0, y ≥ 0 represent 1st quadrant.
Shaded portion XABY is the feasible region, whose vertices are A(6, 0) and B(0, 3).
Here the objective function is Z = 2x + 4y
∴ Z at A(6, 0) = 2(6) + 4(0) = 12
Z at B(0, 3) = 2(0) + 4(3) = 12
∴ Z is minimum at every point along the line segment AB and its minimum value is 12.
Therefore, Z has minimum value at more than two points.
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