Show that the minimum of Z occurs at more than two points. Minimise and Maximise $Z = x + 2y$ subject to $$x + 2y \geq 100, 2x – y \leq 0, 2x + y \leq 200; x, y \geq 0.$$

$\begin{array}{1 1}Min = 100, Max = 400 \\ Min = 250, Max = 400 \\ Min = 100, Max = 250 \\ Min = 0, Max = 100 \end{array}$

Toolbox:
• One we graphically plot the area bounded by the constraints, it’s easy to see which points satisfy all constraints. This common region determined by all the constraints including non-negative constraints of a linear programming problem is called the $\textbf{Feasible Region (or solution region).}$
• Now, any point in the feasible region that gives the optimal value (maximum or minimum) of the objective function is called an $\textbf{Optimal Solution}$. We see that every point in the feasible region satisfies all the constraints, and there are infinitely many points.
• Since we know from theory that the optimal value must occur at a corner point (vertex) of the feasible region, calculate the objective function values associated with the coordinates of all the extreme points. This method is called the $\textbf{Corner Point Method}$.
We are asked to minimize $Z =x+2y \;\;(1)$, subject to the following constraints:
$(2) \quad x+2y \geq 100$
$(3) \quad x+y \geq 60$
$(4) \quad 2x+y \leq 200$
$(5) \quad x \geq 0$
$(6) \quad y \geq 0$
To solve a Linear Programming problem graphically, first plot the constraints for the problem. This is done by plotting the boundary lines of the constraints and identifying the points that will satisfy all the constraints.
$\textbf{Plotting the first constraint}$:
The boundary of the first constraint is represented by the straight line defined by the equation:
$(7) \quad x+2y = 100$
If we can find any two points on this line, the entire line can be plotted by drawing a straight line through these points.
If x = 0, we can see from the equation that y = 50. Thus the point (0,50) must fall on this line. Similarly, if y = 0, we can see from the equation that x = 100. Thus the point (100,0) must fall on this line.
Plot these two points on the graph, and connect them to form the straight line representing the equation (7): x + 2y = 100.
Note that the graph actually extends beyond the x and y axes as shown in the figure. However, we can disregard the points beyond the axes because we have the constraints $x \geq 0, y \geq 0$ and the values assumed by x and y cannot be negative.
The line connecting the points satisfies the equality x+2y = 100. But recall that the first constraint in the problem is the inequality x+2y ≥100
Thus, after plotting the boundary line of a constraint we must determine which area on the graph corresponds to the feasible solutions for original constraint. This can be done very easily by picking an arbitrary point on either side of the boundary and checking where it satisfies the original constraint.
For example, if we test the point (0, 0), we see that it fails the first constraint, i.e., 0 + 0 = 0 which is not ≥ 100. Therefore, the area of the graph must be in the first quadrant, away from the origin.
As shown in the figure, this is the unbounded region with the boundary AB (0,50), (100,0).
$\textbf{Plotting the second constraint}$:
The boundary of the constraint is represented by the straight line defined by the equation: $(8)\; 2x-y= 0$.
If we can find any two points on this line, the entire line can be plotted by drawing a straight line through these points.
If x = 0, we can see from the equation that y = 0 and vice versa. This is the straight line x = y /2 passing through the origin.
As we did with the first constraint, we can disregard the points beyond the axes because we have the constraints x ≥0, y ≥0 and the values assumed by x and y cannot be negative, and we can similarly see that the area bounded by this constraint also lies in the first quadrant.
Substituting (5,0) in the inequality, we get 2*5 – 0 = 10 which is not ≤ 0, so this inequality is towards the Y axis.
$\textbf{Plotting the third constraint}$:
The boundary of the second constraint is represented by the straight line defined by the equation: $(9)\;2x+y = 200$
If we can find any two points on this line, the entire line can be plotted by drawing a straight line through these points.
If x = 0, we can see from the equation that y = 200. If y = 0, we can see that x = 100. Plot these two points (0,200) and (100,0) on the graph, and connect them to form the straight line representing the equation 2x+y = 200.
As we did with the first constraint, we can disregard the points beyond the axes because we have the constraints x ≥0, y ≥0 and the values assumed by x and y cannot be negative, and we can similarly see that the area bounded by this constraint also lies in the first quadrant.
$\textbf{Identifying the Feasible Region}$:
Since there are no other constraints, let us arrive at the feasible region.
Some of the feasible solutions to one constraint in a LP model will usually not satisfy one or more of the other constraints. For example, the point (6,0) satisfies the second constraint but not the first.
Once we draw the graph, it is easy to see which points satisfy all constraints in the model.Notice that when we add the second constraint, some of the feasible solutions associated with the first constraint were eliminated.
We see that the graphs intersect at point E and F. We need to solve the equations (7) and (9) to get the point of intersection (E) and (8) and (9) to arrive at point (F).
To arrive at point E: solve 2x-y = 0 and x+2y = 100.
Therefore, solving, 5y = 200, y = 40. Substituting for y, x = y/2 = 20.
To arrive at point F: Solve 2x-y = 0 and 2x+y = 200.
Solving, we get 4x = 200, x = 50. Substituting, we get y = 100.
Therefore the feasible region is given by the area inside of ACEF as shown in the figure.
$\textbf{Solving the objective function using the Corner Point Method:}$
Recall that any point in the feasible region that gives the optimal value (maximum or minimum) of the objective function is called an optimal solution. We see that every point in the feasible region satisfies all the constraints, and there are infinitely many points.
We know that the solution for an unbounded solution if it exists, will always occur at a point in the feasible region where two or more boundary lines intersect (corner points or vertex).
Since we are asked to minimize the objective function, all we need to do is solve the objective function at the corner points and choose the corner or vertex that gives us the minimize value.
This can be easily summarized in a table as follows:
Corner Point Z = x+2y
A (0,200) Z = 400 (Max Value)
C (0, 50) Z = 100 (Min Value)
E (20,40) Z = 100 (Min Value)
F (50,100) Z = 250

$\textbf{The minimum value of Z is 100 and the maximum is 400. }$