Solve the following L.P.P. by graphical method.Maximize : \(z=4x+6y\)Subject to : \(3x+2y\le12,\ x+y\ge4,\ x,y\ge0\)
Q.6 (A) (ii) Solve the following L.P.P. by graphical method.
Maximize : \(z=4x+6y\)
Subject to : \(3x+2y\le12,\ x+y\ge4,\ x,y\ge0\)
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Best answer
Answer: Maximum \(z=36\) at \(x=0,\ y=6\)
- \(3x+2y=12\) passes through (4, 0) and (0, 6); the region is on the origin side.
- \(x+y=4\) passes through (4, 0) and (0, 4); the region is away from the origin.
[Rough sketch: the feasible region is the triangle in the first quadrant between the two lines, with vertices A (4, 0), B (0, 6) and C (0, 4).]
| Corner point | \(z=4x+6y\) |
|---|---|
| A (4, 0) | 16 |
| B (0, 6) | 36 |
| C (0, 4) | 24 |
The maximum value of z is 36 at B (0, 6).
Doubtly AI
· 28 Oct 2026
AI-verified · Oct 2026
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