Q.Solve the following Linear Programming Problem graphically by using Iso-cost method: Maximize $Z = 3x + 9y$ Subject to the constraints: $x + 4y \leq 8$ $x + 2y \leq 4$ and $x \geq 0, y \geq 0$

Concept: Linear Programming Formulation (Iso-cost / Iso-profit method)

We want to maximise $Z = 3x + 9y$. The constraints are:

$x + 4y \leq 8$

$x + 2y \leq 4$

$x \geq 0, y \geq 0$

Step 1 – Find the feasible region

Plot the lines:

• $x + 4y = 8$ passes through $(8,0)$ and $(0,2)$.
• $x + 2y = 4$ passes through $(4,0)$ and $(0,2)$.

The intersection of the two lines: solving $x + 4y = 8$ and $x + 2y = 4$ gives $2y = 4 \Rightarrow y = 2$, then $x = 0$. So the only intersection is $(0,2)$.

Step 2 – Identify corner points

The feasible region is bounded by $(0,0)$, $(4,0)$, and $(0,2)$. …

We are maximizing a linear function over a polygon. The iso-cost method works because the objective function $Z = 3x + 9y$ is linear — for any fixed value of $Z$, the equation $3x + 9y = c$ is a straight line. As $c$ increases, this line shifts outward (away from the origin). The highest $c$ for which the line still touches the feasible region will hit a corner point of the polygon. That corner is the optimal solution.

Let’s build the feasible region first, then apply the iso-cost method.

1. Plot the constraints as lines.

   Constraint 1: $x + 4y \leq 8$

   The boundary line is $x + 4y = 8$. Find intercepts:

   • If $x = 0$, then $4y = 8 \Rightarrow y = 2$.
   • If $y = 0$, then $x = 8$. So the line passes through $(0, 2)$ and $(8, 0)$. Since the inequality is $\leq$, the feasible side is towards the origin (test $(0,0)$: $0 \leq 8$ is true).

   Constraint 2: $x + 2y \leq 4$

   Boundary: $x + 2y = 4$. Intercepts:

   • $x = 0 \Rightarrow 2y = 4 \Rightarrow y = 2$.
   • $y = 0 \Rightarrow x = 4$. So the line passes through $(0, 2)$ and $(4, 0)$. Again, test $(0,0)$: $0 \leq 4$ is true, so the feasible side is towards the origin.

   Non-negativity: $x \geq 0$, $y \geq 0$ restricts us to the first quadrant.

2. Find the feasible region.

   Both constraints are “less than or equal to” and both include the origin. The feasible region is the intersection of the two half-planes in the first quadrant.

   Notice that both boundary lines pass through $(0, 2)$. Let’s find the other intersection point of the two lines:

   Solve $x + 4y = 8$ and $x + 2y = 4$. Subtract the second from the first:

   $(x + 4y) - (x + 2y) = 8 - 4 \Rightarrow 2y = 4 \Rightarrow y = 2$.

   Then $x + 2(2) = 4 \Rightarrow x = 0$.

   So the two lines intersect at $(0, 2)$ — that’s the only intersection in the first quadrant.

   The feasible region is a triangle with vertices:

   • $(0, 0)$ — origin
   • $(4, 0)$ — from $x + 2y = 4$ on the $x$-axis
   • $(0, 2)$ — intersection point (also on the $y$-axis from both constraints)
   Watch out

   A common mistake is to include $(8,0)$ as a vertex. But $(8,0)$ does not satisfy $x + 2y \leq 4$ (since $8 \leq 4$ is false). So it lies outside the feasible region. Always check every candidate corner against all constraints.

3. Apply the iso-cost method.

   The objective is $Z = 3x + 9y$. Rewrite it as $y = -\frac{1}{3}x + \frac{Z}{9}$. …