Q.30 Solve the following L.P.P. by graphical method :
Maximize : \(z = 10x + 25y\)
Subject to : \(0 \le x \le 3\),
\(0 \le y \le 3\),
\(x + y \le 5\).
Also find the maximum value of \(z\).
Best answer
Answer: \(z\) is maximum at \(x = 2\), \(y = 3\); maximum value \(z = 95\).
Draw the lines \(x = 3\), \(y = 3\) and \(x + y = 5\) (through \((5, 0)\) and \((0, 5)\)). The constraints \(x \ge 0\), \(y \ge 0\), \(x \le 3\), \(y \le 3\), \(x + y \le 5\) hold on the origin side of each line.
[Rough sketch: the feasible region is the pentagon OABCD in the first quadrant, bounded by the axes, \(x = 3\), \(y = 3\) and \(x + y = 5\).]
Corner points: O(0, 0), A(3, 0), B(3, 2) (intersection of \(x = 3\) and \(x + y = 5\)), C(2, 3) (intersection of \(y = 3\) and \(x + y = 5\)), D(0, 3).
| Corner point | \(z = 10x + 25y\) |
|---|---|
| O(0, 0) | 0 |
| A(3, 0) | 30 |
| B(3, 2) | 30 + 50 = 80 |
| C(2, 3) | 20 + 75 = 95 |
| D(0, 3) | 75 |
The maximum value of \(z\) is 95, at \(x = 2\), \(y = 3\).
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