Corner points: (0,8) gives z=560; (10,0) gives z=500; intersection of 2x+y=8 and x+2y=10 is (2,4) giving z=100+280=380. Minimum is 380.
Consider the linear programming problem : Minimize z = 50x + 70y Subject to 2x+y ≥ 8, x+2y ≥ 10, x ≥ 0, y ≥ 0 The minimum value of objective function…
CUET UG 2022 — Mathematics Algebra
Consider the linear programming problem :
Minimize z=50x+70y
Subject to 2x+y≥8, x+2y≥10, x≥0, y≥0
The minimum value of objective function is :
Held on 17 Aug 2022 · Verified 13 Jul 2026.
Options
- A
380
- B
560
- C
500
- D
120
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