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The minimum value of Z=30x+10yZ = 30x + 10y subject to the constraints x+2y≤30,3x+y≥30,4x+3y≥60,x,y≥0x + 2y \leq 30, 3x + y \geq 30, 4x + 3y \geq 60, x, y \geq 0 is

Solution

✅ Correct Option: 3

Feasible corners: (15,0),(30,0),(6,12)(15, 0), (30, 0), (6, 12). Evaluating Z=30x+10yZ = 30x + 10y:

(15,0):450(15, 0): 450

(30,0):900(30, 0): 900

(6,12):180+120=300(6, 12): 180 + 120 = 300.

Minimum Z=300Z = 300 at (6,12)(6, 12).

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