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The maximum value of Z=3x+yZ = 3x + y subject to the constraints x+y≤30,2x+y≤40,x,y≥0x + y \leq 30, 2x + y \leq 40, x, y \geq 0 is

Solution

✅ Correct Option: 4

Corner points: (0,0),(20,0),(10,20),(0,30)(0,0), (20,0), (10,20), (0,30). Evaluating Z=3x+yZ = 3x + y:

(0,0):0(0,0): 0

(20,0):60(20,0): 60

(10,20):50(10,20): 50

(0,30):30(0,30): 30.

Maximum Z=60Z = 60 at (20,0)(20, 0).

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