The maximum value of the objective function $z = 2x + 3y$ of an L.P.P. subjected to the constraints $x-y≤1,x + y ≤3,x,y≥0$, is

9

The correct answer is Option (2) → 9

Given:

Objective function: $z = 2x + 3y$

Subject to constraints:

• $x - y \leq 1$
• $x + y \leq 3$
• $x \geq 0,\ y \geq 0$

Step 1: Find corner points of feasible region

Intersection of $x - y = 1$ and $x + y = 3$:

Solving:

• Add: $2x = 4 \Rightarrow x = 2$
• Substitute: $2 + y = 3 \Rightarrow y = 1$

⇒ Point A: $(2, 1)$

Intersection with axes:

• $x - y = 1$ and $y = 0$ ⇒ $x = 1$ ⇒ Point B: $(1, 0)$
• $x + y = 3$ and $x = 0$ ⇒ $y = 3$ ⇒ Point C: $(0, 3)$

Also include origin $(0, 0)$

Step 2: Evaluate $z = 2x + 3y$ at all corner points

• At $(0, 0)$: $z = 0$
• At $(1, 0)$: $z = 2$
• At $(2, 1)$: $z = 2(2) + 3(1) = 4 + 3 = 7$
• At $(0, 3)$: $z = 9$