Q.30 Solve the L.P.P. graphically :
Minimize : \(z = 5x + 2y\),
Subject to, \(5x + y \ge 10,\ x + y \ge 6,\)
\(x \ge 0,\ y \ge 0\)
Best answer
Answer: Minimum \(z = 15\) at \(x = 1,\ y = 5\).
Convert the constraints to equations and find the intercepts:
| Line | Points | Region |
|---|---|---|
| \(5x + y = 10\) | \((2, 0),\ (0, 10)\) | away from origin (origin side fails: \(0 \ge 10\) false) |
| \(x + y = 6\) | \((6, 0),\ (0, 6)\) | away from origin |
[Figure: rough sketch in the first quadrant with the lines \(5x + y = 10\) through (2, 0) and (0, 10), and \(x + y = 6\) through (6, 0) and (0, 6). The feasible region is the unbounded region above both lines, with corner points A(6, 0), B(1, 5) and C(0, 10).]
Intersection of \(5x + y = 10\) and \(x + y = 6\): subtracting, \(4x = 4\), so \(x = 1,\ y = 5\). Point B(1, 5).
| Corner point | \(z = 5x + 2y\) |
|---|---|
| A(6, 0) | 30 |
| B(1, 5) | 15 |
| C(0, 10) | 20 |
The smallest value is 15 at B(1, 5). Although the region is unbounded, the half-plane \(5x + 2y \lt 15\) has no point in common with the feasible region, so this is the minimum.
Hence \(z\) is minimum at \(x = 1,\ y = 5\), and \(z_{\min} = 15\).
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