For LPP: Maximize $z = 2x + 3y$ subject to the constraints

$x + y \geq 2$, $x + 2y \geq 3$, $x \geq 0$, $y \geq 0$,

which of the following graph represents the feasible region of the above LPP as shaded portion?

The feasible region is the area where all constraints are satisfied.

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For the constraint $x + y \geq 2$, first draw the line $x + y = 2$.

When $x = 0$, $y = 2$ giving point $(0, 2)$

When $y = 0$, $x = 2$ giving point $(2, 0)$

Since the constraint is $\geq$, the region is above the line.

Testing point $(0, 0)$: $0 + 0 = 0 \not\geq 2$, so $(0, 0)$ is not in the region.

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For the constraint $x + 2y \geq 3$, draw the line $x + 2y = 3$.

When $x = 0$, $2y = 3$

$y = 1.5$ giving point $(0, 1.5)$

When $y = 0$, $x = 3$ giving point $(3, 0)$

Since the constraint is $\geq$, the region is above the line.

Testing point $(0, 0)$: $0 + 0 = 0 \not\geq 3$, so $(0, 0)$ is not in the region.

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The constraints $x \geq 0$ and $y \geq 0$ restrict the region to the first quadrant.

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The feasible region is where all constraints are satisfied simultaneously. Both main constraints have $\geq$ signs, so the region is above both lines and in the first quadrant. This creates an unbounded region extending infinitely upward and rightward.

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The intersection of the two lines $x + y = 2$ and $x + 2y = 3$ gives a key point.

From the first equation: $x = 2 - y$

Substituting into the second equation:

$(2 - y) + 2y = 3$

$2 + y = 3$

$y = 1$

$x = 2 - 1 = 1$

The intersection point is $(1, 1)$.

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The feasible region is:

• In the first quadrant only
• Above both constraint lines
• Unbounded (extends to infinity)
• Has corner points at $(0, 2)$ and $(1, 1)$

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Option 4 is correct because it shows an unbounded region extending upward and rightward, correctly shades the area above both constraint lines, and stays in the first quadrant.

Other options likely show bounded regions or shade the wrong area below the lines instead of above.