The maximum value of Z = 3x + 4y subjected to contraints x + y ≤ 40, x + 2y ≤ 60, x ≥ 0 and y ≥ 0 is

The correct option is 140.

To find the maximum value of the objective function
$$Z=3x+4y$$
subject to the given constraints, we can use the graphical method of linear programming. The constraints are:

1.
$x+y\le 40$

2.
$x+2y\le 60$

3.
$x\ge 0$
and
$y\ge 0$

Step 1: Find the boundary lines and their intercepts
Let us convert the inequalities into equations to find the boundary lines of the feasible region.
For the first line, $x+y=40$:
- If $x=0$, then $y=40$. The point is $(0,40)$.
- If $y=0$, then $x=40$. The point is $(40,0)$.

For the second line, $x+2y=60$:
- If $x=0$, then $y=30$. The point is $(0,30)$.
- If $y=0$, then $x=60$. The point is $(60,0)$.

Step 2: Find the point of intersection of the two lines
We solve the system of equations:
1)
$x+y=40$
2)
$x+2y=60$
Subtracting the first equation from the second equation:
$$(x+2y)-(x+y)=60-40$$
$$y=20$$
Substituting $y=20$ back into $x+y=40$ gives:
$$x+20=40\Rightarrow x=20$$
The point of intersection is $(20,20)$.

Step 3: Identify the corner points of the feasible region
Since $x\ge 0$ and $y\ge 0$, the feasible region lies in the first quadrant. By testing the origin $(0,0)$ in the inequalities, we find that it satisfies both constraints. The corner points of the bounded feasible region are:
- $O(0,0)$
- $A(40,0)$
- $B(20,20)$
- $C(0,30)$

Step 4: Evaluate the objective function Z at each corner point
- At $O(0,0)$:
$$Z=3\left(0\right)+4\left(0\right)=0$$
- At $A(40,0)$:
$$Z=3\left(40\right)+4\left(0\right)=120$$
- At $B(20,20)$:
$$Z=3\left(20\right)+4\left(20\right)=60+80=140$$
- At $C(0,30)$:
$$Z=3\left(0\right)+4\left(30\right)=120$$

Comparing the values, the maximum value of $Z$ is $140$, which occurs at the corner point $(20,20)$.