A linear programming problem is as follows:

Minimize $z = 2x + 3y$

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

(7, 7) and (3, 3)

The correct answer is Option (3) → (7, 7) and (3, 3)

Given LPP:

Minimize: $z = 2x + 3y$

Subject to constraints:

• $x \geq 3$
• $x \leq 9$
• $y \geq 0$
• $x - y \geq 0 \Rightarrow y \leq x$
• $x + y \leq 14$

To find corner points, graph all lines and find the intersection points within the feasible region.

Lines involved:

1. $x = 3$
2. $x = 9$
3. $y = 0$
4. $y = x$
5. $x + y = 14$

Now calculate intersection points:

1. Intersection of $x = 3$ and $y = 0$: (3, 0)
2. Intersection of $x = 3$ and $x + y = 14$: $y = 11$ → (3, 11) → not feasible (since $y > x$)
3. Intersection of $x = 3$ and $y = x$: $y = 3$ → (3, 3)
4. Intersection of $x = 9$ and $y = 0$: (9, 0)
5. Intersection of $x = 9$ and $x + y = 14$: $y = 5$ → (9, 5)
6. Intersection of $y = x$ and $x + y = 14$:

Let $y = x$ → $x + x = 14 \Rightarrow x = 7$, $y = 7$ → (7, 7)

Now test feasibility for all points:

• (3, 0): ✔️
• (3, 3): ✔️
• (7, 7): ✔️
• (9, 0): ✔️
• (9, 5): ✔️

Therefore, the feasible region has 5 corner points:

${(3, 0), (3, 3), (7, 7), (9, 5), (9, 0)}$