Q.Solve the following Linear Programming Problem graphically by using Iso-cost method: Maximize $Z = 3x + 2y$ Subject to the constraints: $-2x + y \leq 1$ $x + y \leq 3$ $x \leq 2$ and $x \geq 0, y \geq 0$

Concept: Linear Programming Formulation — Iso-cost method (graphical).

Step 1 – Plot constraints

Convert each inequality to an equation and draw the lines:

• $-2x + y = 1$ passes through $(0,1)$ and $(-0.5,0)$
• $x + y = 3$ passes through $(0,3)$ and $(3,0)$
• $x = 2$ is a vertical line Feasible region is the intersection of half-planes with $x, y \geq 0$.

Step 2 – Identify corner points

The feasible region is bounded. Corner points:

• $A(0,0)$
• $B(0,1)$ (from $-2x + y = 1$ and $y$-axis)
• $C(2,0)$ (from $x=2$ and $x$-axis)
• $D(2,1)$ (intersection of $x=2$ and $x+y=3$)
• $E\left(\tfrac{2}{3},\tfrac{7}{3}\right)$ (intersection of $-2x+y=1$ and $x+y=3$) …

Why the Iso-Cost Method Works

In linear programming, the objective function $Z = 3x + 2y$ represents a family of parallel lines — each line corresponds to a different value of $Z$. As we slide this line outward (increasing $Z$), the last point it touches in the feasible region gives the optimal solution. This is the iso-cost (or iso-profit) method: we draw one such line for a convenient $Z$, then shift it parallel until it just leaves the feasible region.

The key insight: because the feasible region is convex and the objective is linear, the optimum always occurs at a corner point (vertex) of the region. So we can either slide the line visually or simply evaluate $Z$ at all vertices.

Step-by-Step Solution

1. Convert inequalities to equations and plot the lines

We have four constraints (including non-negativity):

• $-2x + y = 1$
• $x + y = 3$
• $x = 2$
• $x = 0$ and $y = 0$ (axes)

Let’s find intercepts for each line:

Since $x \geq 0, y \geq 0$, we only care about the first quadrant.

2. Determine the feasible region

We shade the half-planes that satisfy each inequality:

• For $-2x + y \leq 1$: test $(0,0)$ gives $0 \leq 1$ (true), so shade below the line $-2x + y = 1$.
• For $x + y \leq 3$: test $(0,0)$ gives $0 \leq 3$ (true), so shade below $x + y = 3$.
• For $x \leq 2$: shade left of $x = 2$.
• $x \geq 0, y \geq 0$ restricts to the first quadrant.

The feasible region is the polygon bounded by these lines and axes.

3. Find the corner points (vertices)

The vertices are intersections of boundary lines:

A: Intersection of $x=0$ and $y=0$ → $(0,0)$

B: Intersection of $y=0$ and $x + y = 3$ → $(3,0)$

But check $x \leq 2$: $3 \leq 2$ is false. So $(3,0)$ is not feasible.

Instead, intersection of $y=0$ and $x=2$ → $(2,0)$

C: Intersection of $x=2$ and $x + y = 3$ → $2 + y = 3 \Rightarrow y = 1$ → $(2,1)$

D: Intersection of $x + y = 3$ and $-2x + y = 1$

Subtract second from first: $(x+y) - (-2x+y) = 3 - 1 \Rightarrow 3x = 2 \Rightarrow x = \frac{2}{3}$

Then $y = 3 - x = 3 - \frac{2}{3} = \frac{7}{3}$ → $\left(\frac{2}{3}, \frac{7}{3}\right)$

E: Intersection of $-2x + y = 1$ and $x=0$ → $y = 1$ → $(0,1)$

F: Intersection of $x=0$ and $x + y = 3$ → $(0,3)$

But check $-2(0) + 3 \leq 1$? $3 \leq 1$ is false. So $(0,3)$ is not feasible. …