CMA Final · Strategic Cost Management · Linear Programming
Kaveri Ltd maximises Z = 5x + 4y subject to 6x + 4y ≤ 24 and x + 2y ≤ 6, x, y ≥ 0. What is the optimal value of Z?
The optimal value is Rs 21. The two constraints intersect at x = 3 and y = 1.5, giving Z = 15 + 6 = 21, which exceeds the other corner values of 20 at (4,0) and 12 at (0,3).
- ARs 20
- BRs 21Correct
- CRs 24
- DRs 22
Explanation
Intersection: from 6x+4y=24 and x+2y=6, double the second: 2x+4y=12; subtract: 4x=12, x=3, y=1.5. Z=15+6=21. Other corners: (4,0)=20, (0,3)=12. So 21 is the maximum; 20 is the axis corner only.
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