Q.30 Solve the following L.P.P. using graphical method :
Maximize, \(z = 9x + 13y\)
Subject to, \(2x + 3y \le 18\),
\(2x + y \le 10\)
\(x \ge 0,\ y \ge 0\)
Best answer
Answer: Maximum \(z = 79\) at \(x = 3,\ y = 4\).
Draw the lines \(2x + 3y = 18\) (through (9, 0) and (0, 6)) and \(2x + y = 10\) (through (5, 0) and (0, 10)). Both inequalities are of the \(\le\) type, so the origin side is feasible; with \(x, y \ge 0\) the feasible region is the quadrilateral OABC.
[Graph: feasible region OABC bounded by the axes and the two lines, with vertices O(0, 0), A(5, 0), B(3, 4), C(0, 6).]
Point B: solving \(2x + 3y = 18\) and \(2x + y = 10\) gives \(2y = 8\), \(y = 4\), \(x = 3\).
| Vertex | \(z = 9x + 13y\) |
|---|---|
| O(0, 0) | 0 |
| A(5, 0) | 45 |
| B(3, 4) | 27 + 52 = 79 |
| C(0, 6) | 78 |
\(z\) is maximum at B(3, 4). Maximum value of \(z = 79\) when \(x = 3,\ y = 4\).
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