Q.Solve the following Linear Programming Problem Graphically. Maximize $Z = 6x + y$ Subject to constraints: $2x + y \geq 3$ $y - x \geq 0$ $x \geq 0$, and $y \geq 0$

Concept: Linear Programming – Graphical Method

The feasible region is determined by plotting all constraints and finding their intersection. Since this is a maximization problem with "greater than or equal to" constraints (except non-negativity), we need to identify the unbounded feasible region and check whether the objective function has a finite maximum.

Step 1: Plot the constraint lines.

• $2x + y = 3$ passes through $(0, 3)$ and $(1.5, 0)$; the region is above this line.
• $y = x$ passes through the origin with slope 1; the region is above this line (where $y \geq x$).
• Combined with $x \geq 0, y \geq 0$, we work in the first quadrant.

Step 2: Find corner points of the feasible region.

The binding constraints intersect where $2x + y = 3$ and $y = x$:

So the corner point is $(1, 1)$.

The line $y = x$ intersects the $y$-axis at $(0, 0)$, but $(0,0)$ does not satisfy $2x + y \geq 3$. The line $2x + y = 3$ meets the $y$-axis at $(0, 3)$. …

Why the graphical method works

Linear programming finds the best value of a linear objective function over a region defined by linear inequalities. The feasible region is a convex polygon (or unbounded convex set), and a fundamental theorem tells us that if an optimum exists, it occurs at a corner point. We plot the constraints, identify the feasible region, evaluate $Z$ at each vertex, and check whether the region allows $Z$ to grow without bound.

Step-by-step solution

1. Rewrite the constraints in a form easy to graph

We have:

• $2x + y \geq 3 \implies y \geq 3 - 2x$
• $y - x \geq 0 \implies y \geq x$
• $x \geq 0$, $y \geq 0$ (first quadrant)

Each inequality defines a half-plane; the feasible region is their intersection.

2. Plot the boundary lines

For $2x + y = 3$:

• When $x = 0$, $y = 3$ → point $(0, 3)$
• When $y = 0$, $x = 1.5$ → point $(1.5, 0)$

For $y = x$:

• A line through the origin with slope $1$, passing through $(0,0)$, $(1,1)$, $(2,2)$, etc.

3. Identify the feasible region

We need:

• Points above or on the line $y = 3 - 2x$
• Points above or on the line $y = x$
• Points in the first quadrant

The line $y = x$ and $y = 3 - 2x$ intersect where:

So the intersection point is $(1, 1)$.

Now check which region satisfies all constraints:

• The line $y = 3 - 2x$ has a negative slope; we want the region above it.
• The line $y = x$ divides the first quadrant; we want the region above it (where $y \geq x$).

The feasible region is the area in the first quadrant that lies above both lines. The corner points are:

• $(0, 3)$ — intersection of $x = 0$ and $2x + y = 3$
• $(1, 1)$ — intersection of $y = x$ and $2x + y = 3$
• The region extends infinitely along the ray $y = x$ for $x \geq 1$

4. Evaluate the objective function at corner points

At $(0, 3)$:

At $(1, 1)$:

5. Check the direction of increase of $Z = 6x + y$

The objective function can be rewritten as:

$y = -6x + Z$ …