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A linear programming problem is as follows:

Minimize z=2x+3yz = 2x + 3y

Subject to the constraints x≥3,x≤9,y≥0,x−y≥0,x+y≤14x \ge 3, x \le 9, y \ge 0, x - y \ge 0, x + y \le 14. The feasible region has 5 corner points including

Solution

✅ Correct Option: 3

The constraints are:

x≥3x \ge 3, x≤9x \le 9, y≥0y \ge 0, x−y≥0x - y \ge 0 (i.e. y≤xy \le x), x+y≤14x + y \le 14

Corner points are formed where two boundary lines intersect inside the feasible region.


x=3x = 3 meets y=0y = 0

Corner point =(3,  0)= (3,\; 0)


x=9x = 9 meets y=0y = 0

Corner point =(9,  0)= (9,\; 0)


x=9x = 9 meets x+y=14x + y = 14

y=14−9=5y = 14 - 9 = 5

Corner point =(9,  5)= (9,\; 5)


x+y=14x + y = 14 meets y=xy = x

x+x=14x + x = 14

2x=142x = 14

x=7,  y=7x = 7,\; y = 7

Corner point =(7,  7)= (7,\; 7)


x=3x = 3 meets y=xy = x

y=3y = 3

Corner point =(3,  3)= (3,\; 3)


Checking option 1: (0,0)(0, 0) fails x≥3x \ge 3 — not in feasible region. ❌

Checking option 2: (14,0)(14, 0) fails x≤9x \le 9 — not in feasible region. ❌

Checking option 3: (7,7)(7, 7) and (3,3)(3, 3) — both are valid corner points. ✅

Checking option 4: At (3,6)(3, 6), x−y=3−6=−3<0x - y = 3 - 6 = -3 < 0 — fails x−y≥0x - y \ge 0. ❌


The answer is Option 3 — (7,7)(7, 7) and (3,3)(3, 3).

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