It seems like the inequality symbols in the constraints are missing. From the context, I will assume the constraints should look something like this:

$\text{Minimize } Z = 6x + 2y$ $\text{Subject to:}$

1. $x + 2y \geq 30$
2. $x + 2y \leq 120$
3. $2x + y \leq 40$
4. $x \geq 0$, $y \geq 0$

Let's proceed to solve this Linear Programming Problem (LPP) graphically.

Steps to solve graphically:

1. Plot the lines for the constraints:

   • For $x + 2y = 30$ and $x + 2y = 120$, find the points of intersection with the axes.
   • For $2x + y = 40$, find the points of intersection with the axes.

2. Shade the feasible region:

   • Identify the region that satisfies all the inequalities.

3. Identify the corner points:

   • The feasible region will have corner points where the lines intersect.

4. Evaluate the objective function:

   • Substitute the corner points into $Z = 6x + 2y$ to find the minimum value.

I'll walk you through the solution and plot it graphically. Let's begin by solving for the points where the lines intersect the axes.

Step 1: Find the intercepts

For $x + 2y = 30$:

• When $x = 0$, $2y = 30$ → $y = 15$.
• When $y = 0$, $x = 30$.

Thus, the intercepts are $(30, 0)$ and $(0, 15)$.

For $x + 2y = 120$:

• When $x = 0$, $2y = 120$ → $y = 60$.
• When $y = 0$, $x = 120$.

Thus, the intercepts are $(120, 0)$ and $(0, 60)$.

For $2x + y = 40$:

• When $x = 0$, $y = 40$.
• When $y = 0$, $2x = 40$ → $x = 20$.

Thus, the intercepts are $(20, 0)$ and $(0, 40)$.

Step 2: Plot the lines and identify the feasible region

Now we will plot these lines on a graph and find the feasible region that satisfies all the constraints.

Let me plot this graphically.