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The point/points which provides the optimal solution of the linear programming problem maximum z=3x+yz = 3x+y, x+3y≤12,3x+y≤12,x,y≥0x+3y \le 12, 3x+y \le 12, x,y \ge 0, is/are

Solution

✅ Correct Option: 3

Corner points are (0,0)(0,0), (4,0)(4,0), (3,3)(3,3) and (0,4)(0,4), with z=0,12,12,4z=0,12,12,4.

The maximum 12 occurs at two adjacent corners (4,0)(4,0) and (3,3)(3,3), both lying on 3x+y=123x+y=12, so every point of that segment is optimal.

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