Solve the following Linear Programming Problems graphically:

Minimize: Z = x + 2y

Subjected to:

2x + y ≥ 3
x + 2y ≥ 6
x, y ≥ 0

To solve the given Linear Programming Problem (LPP) graphically, we need to minimize the objective function Z = x + 2y, subject to the following constraints:

2x + y ≥ 3
x + 2y ≥ 6
x ≥ 0, y ≥ 0
Let's break it down step by step:

Step 1: Plot the constraints
We need to convert each constraint into an equation to plot the lines on the graph.

Constraint 1: 2x + y ≥ 3
Rearrange the inequality as an equation: 2x + y = 3 Solve for y: y = 3 - 2x

This is a straight line with slope -2 and y-intercept 3.

Constraint 2: x + 2y ≥ 6
Rearrange the inequality as an equation: x + 2y = 6 Solve for y: y = (6 - x) / 2

This is a straight line with slope -1/2 and y-intercept 3.

Constraint 3: x ≥ 0, y ≥ 0
This means that the feasible region is in the first quadrant of the coordinate plane.

Step 2: Graph the constraints
Plot the lines 2x + y = 3 and x + 2y = 6 on a graph.
The feasible region is the area where all constraints are satisfied simultaneously.
Since both x and y are non-negative, the feasible region will be in the first quadrant.
Step 3: Identify the corner points (vertices) of the feasible region
The corner points of the feasible region are where the constraint lines intersect. We need to find the points of intersection.

Intersection of 2x + y = 3 and x + 2y = 6
Solve the system of equations:

2x + y = 3
x + 2y = 6
Multiply the first equation by 2 to make the coefficients of y the same: 4x + 2y = 6 Now subtract the second equation from this: (4x + 2y) - (x + 2y) = 6 - 6 3x = 0 x = 0

Substitute x = 0 into 2x + y = 3: 2(0) + y = 3 y = 3

So, the intersection point is (0, 3).

Intersection of 2x + y = 3 and x = 0
Substitute x = 0 into 2x + y = 3: 2(0) + y = 3 y = 3

So, the point is (0, 3), which is the same as the previous intersection point.

Intersection of x + 2y = 6 and x = 0
Substitute x = 0 into x + 2y = 6: 0 + 2y = 6 y = 3

So, the point is (0, 3), which is the same as before.

Intersection of x + 2y = 6 and y = 0
Substitute y = 0 into x + 2y = 6: x + 2(0) = 6 x = 6

So, the intersection point is (6, 0).

Step 4: Evaluate the objective function Z = x + 2y at the corner points
Now, we evaluate the objective function at the corner points (0, 3) and (6, 0).

At (0, 3): Z = 0 + 2(3) = 6

At (6, 0): Z = 6 + 2(0) = 6

Step 5: Find the minimum value of Z
The objective function Z is minimized when Z = 6 at both (0, 3) and (6, 0).

Final Answer:
The minimum value of Z = x + 2y is 6, and it occurs at the points (0, 3) and (6, 0).