The shaded region is the feasible region ABC determined by the system of constraints (2) to (4), which is bounded. The coordinates of corner points

A, B and C are (0,5), (4,3) and (0,6) respectively. Now we evaluate Z = 200x + 500y at these points.
Hence, minimum value of Z is 2300 attained at the point (4, 3)

The system of constraints is:
\(x-y\le 1\) ..(1)
\(x+y\ge 3\)...(2)
and \(x\ge 0,y\ge 0\)...(3)
The shaded region in the following figure is the feasible region determined by (1)-(3).

It is observed that feasible region is not bounded.
Hence, the max value of Z is not finite.

The feasible region determined by the constraints, x + y ≤ 4, x ≥ 0, y ≥ 0, is as follows.
The corner points of the feasible region are O (0, 0), A (4, 0), and B (0, 4). The values of Z at these points are as follows.

Therefore, the maximum value of Z is 16 at the point B (0, 4).

Our problem is to maximise Z = - x + 2Y  ...(i)
subject to constraints,
\(x \geq 3\) ...(ii)
\(x+y \geq 5\)  ...(iii)
\(x+2 y \geq 6\)  ...(iv)
Table for line x + y = 5 is
\(\begin{array}{c|c|c} \hline x & 0 & 5 \\ \hline y & 5 & 0 \\ \hline \end{array}\) 
So, the line passes through the points (0, 5) and (5, 0).
On putting (0, 0) in the inequality \(x+y \geq 5\), we get
\(0+0 \geq 5\) 
\(\Rightarrow 0 \geq 5\), which is not true.
So, the half plane is away from the origin .
Table for line \(x+2 y=6\) is
 \(\begin{array}{c|c|c} \hline x & 0 & 6 \\ \hline y & 3 & 0 \\ \hline \end{array}\) 
So, the line passes through the points (0,3) and (6,0). 
On putting (0, 0) in the inequality \(x+2 y \geq 6\), we get
\(0+2 \times 0 \geq 6\) 
\(\Rightarrow 0 \geq 6\) which is not true.
So, the half plane is away from the origin
Also,\(x \geq 3\), so the region is away from the orgin. 
Since \(x \geq 0, y \geq 0\) 
So, the region lies in the I quadrant. The points of intersection of lines x = 3 and x + y = 5 is C(3, 2) and lines x + 2y = 6 and x + y = 5 is B(4, 1). It can be seen that the feasible region is unbounded.

The corner points of the feasible region are A(6, 0), B(4, 1) and C(3, 2).
The values of Z at corner points are given below
\(\begin{array}{c|c} \hline \text { Corner points } & z=-x+2 y \\ \hline A(6,0) & Z=-6+2 \times 0=-6 \\ B(4,1) & Z=-4+2 \times 1=-2 \\ C(3,2) & Z=-3+2 \times 2=1 \\ \hline \end{array}\) 
As the feasible region is unbounded, therefore Z = 1 may or may not be the maximum value. For this, we draw the graph of the inequality \(-x+2 y>1\) and check whether the resulting half plane has points in common with the feasible region or not. The feasible region have points in common with the \(-x+2 y>1.\) Therefore, Z = 1 is not the maximum value.
Hence, Z has no maximum value.

First of all, let us graph the feasible region of the system of inequalities (2) to (5). The feasible region (shaded). Observe that the feasible region is unbounded.
We now evaluate Z at the corner points.

From this table, we find that – 300 is the smallest value of Z at the corner point (6, 0). Can we say that minimum value of Z is – 300? Note that if the region would have been bounded, this smallest value of Z is the minimum value of Z (Theorem 2). But here we see that the feasible region is unbounded. Therefore, – 300 may or may not be the minimum value of Z. To decide this issue, we graph the inequality
– 50x + 20y < – 300 (see Step 3(ii) of corner Point Method.)
i.e., – 5x + 2y < – 30
and check whether the resulting open half plane has points in common with feasible region or not. If it has common points, then –300 will not be the minimum value of Z.
Otherwise, –300 will be the minimum value of Z. 
It has common points. Therefore, Z = –50 x + 20 y has no minimum value subject to the given constraints.
In the above example, can you say whether z = – 50 x + 20 y has the maximum value 100 at (0, 5)? For this, check whether the graph of – 50 x + 20 y > 100 has points in common with the feasible region.

Our problem is to minimise Z = x + 2y   ...(i)
subject to constraints,
\(2 x+y \geq 3\) ...(ii)
\(x+2 y \geq 6\)   ...(iii)
and \(x \geq 0, y \geq 0\)   ...(iv)
Table for line 2x + y = 3 is
\(x \geq 0, y \geq 0\) 
Table for line 2x + y = 3 is
\(\begin{array}{c|c|c} \hline x & 0 & 3 / 2 \\ \hline y & 3 & 0 \\ \hline \end{array}\) 
So, line passes through the points (0, 3) and \(\left(\frac{3}{2}, 0\right)\) .
On putting (0, 0) in the inequality \(2 x+y \geq 3\) we get
\(2 \times 0+0 \geq 3\) 
\(\Rightarrow 0 \geq 3\), which is not true
So, the half plane is away from the origin
Table for line x + 2y = 6 is
\(\begin{array}{c|c|c} \hline x & 0 & 6 \\ \hline y & 3 & 0 \\ \hline \end{array}\) 
So, line passes through the points (0, 3) and (6, 0). 
On putting (0, 0) in the inequality \(x+2 y \geq 6\) ,we get
\(0+2 \times 0 \geq 6\) 
\(\Rightarrow 0 \geq 6\), which is not true.
So, the half plane is away from the origin.
Also,\(x \geq 0 \text { and } y \geq 0\), so the region lies in the I quadrant
The intersection point of the lines x + 2y = 6 and \(2 x+y=3\) B(0, 3).
 
The corner points of the feasible region are A(6, 0) and B(0, 3).
The values of Z at the corner points are given below
As the feasible region is unbounded, therefore 6 mayor may not be the minimum value of Z. For this we draw a dotted graph of the inequality x + 2y.
< 6 and check whether the resulting half plane has points in common with a feasible region or not.
It can be seen that the feasible region has no common point with x + 2y < 6. Therefore, the minimum value of Z is 6. Thus,
the minimum value of Z occurs at more than 2 points. [The value of Z is minimum at every point on the line segment AB] 
Let \(f(x, y)=x+2 y-6\)
At point \(P(5,2), f(5,2)=5+2 \times 2-6=3>0\) 
Hence, point lies inside the shaded region.
Now, distance between 0 and P
\(=\sqrt{(5-0)^{2}+(2-0)^{2}}=\sqrt{25+4}=\sqrt{29}=5.39\)

Given inequations are
\(2 x+4 y \leq 8 \text { or } \ x+2 y \leq 4 \)
\(3 x+y \leq 6, x+y \leq 4, \ x \geq 0, \ y \geq 0 \)
Maximise Z = 2x + 5y on plotting the graph of the inequations we notice shaded portion as feasible solution

Possible points for maximum Z are A(2, 0),\(B\left(\frac{8}{5}, \frac{6}{5}\right)\) C (0,2)

Z is maximum at qo, 2), i.e. x = 0, y = 2, maximum value = 10

We have, maximise, Z = 60x + 40y             ...(i)
Subject to the constraints, x + 2y \(\leq 12\)     ...(ii)
\(\begin{aligned} 2 x+y & \leq 12 \end{aligned}\)                                   ...(iii)
\(\begin{aligned} 4 x+5 y & \geq 20 \end{aligned}\)                                 ...(iv)
and \(\begin{aligned} x, y & \geq 0 \end{aligned}\)                                     ...(v)
Table for  line x + 2y = 12 is

So, the line x + 2y = 12 is passing through the points (0, 6) and (12, 0).
On putting (0,0) in the inequality \(x+2 y \leq 12\) we get \(0+2(0) \leq 12 \Rightarrow 0 \leq 12\), which is true
So,the half plane is towards the origin.
Table for line 2x + y = 12 is 

So, the line 2x + y = 12 is passing through the points (0, 12) and (6, 0).
On putting (0, 0) in the inequality \(2 x+y \leq 12\) we get
\(2(0)+0 \leq 12 \Rightarrow 0 \leq 12\) , which is true.
So, the half plane is towards the origin.
Table for line 4x + 5y = 20 is

So, the line 4x + 5y = 20 is passing through the points (0,4) and (5, 0).
On putting (0, 0) in the inequality \(4 x+5 y \geq 20\), we get 4(0) + 5(0) \(\geq\) 20 \(\Rightarrow\) 0 > 20 which is not true.
So, the half plane is away from the origin.
Also, x, y \(\geq\) 0
So, the region lies in Ist quadrant.

On solving Eqs. x + 2y = 12 and 2x + y = 12, we get D(4, 4)
Clearly, the feasible region is ABCDEA.
The corner points of the feasible region are A(0, 4), B(5, 0), C(6, 0), D(4, 4) and E(0, 6).
The value of Z at corner points are given below.

The maximum value of Z is 400 at D(4, 4).

Given that,
Minimise and Maximise Z = 3x + 9y       ...(i)
Subject to the constraints are
\(\begin{aligned} x+3 y & \leq 60 \end{aligned}\)        ...(ii)
\(\begin{aligned} x+y & \geq 10 \end{aligned}\)          ...(iii)
\(\begin{aligned} x & \leq y \end{aligned}\)                  ...(iv)
\(\begin{aligned} x \geq 0, y & \geq 0 \end{aligned}\)         ...(v)

First of all,let us plot the graph of the feasible region of the system of linear inequalities (ii) to (v). The feasible region ABCDA is shown in the figure.
Note That the region is bounded. The coordinates of the corner points A, B, C and D are (0, 10), (5, 5), (15, 15) and (0, 20), respectively.

We, now find the minimum and maximum value of Z. From the table, we find that the minimum value of Z is 60 at the point B(5, 5) of the feasible region.
The maximum value of Z on the feasible region occurs at the two corner points C (15, 15) and D (0, 20) and it is 180 in each case.
Remark Observe that in the above example, the problem has multiple optimal solutions at the corner points C and D, i.e. the both points produce same maximum value 180.
In such cases, you can see that every point on the line segment CD joining the two corner points C and D also give the same maximum value. Same is also true in the case, if the two points produce same minimum value.

Our problem is to maximise
Z = 3x + 4y                 ...(i)
Subject to the constraints are
\(\begin{aligned} x+y & \leq 4 \end{aligned}\)           ...(ii)
\(\begin{aligned} x & \geq 0, y \geq 0 \end{aligned}\)       ...(iii)
Table for the line x + y = 4 is

On putting (0, 0) in the inequality x+ y \(\leq\) 4, we have
0 + 0 \(\leq\) 4
\(\Rightarrow\)    0 \(\leq\) 4 (which is true)
So, the half plane is towards the origin

So, the feasible region lies in the Ist quadrant.
\(\therefore\) Feasible region is OABO.
The corner points of the feasible region are O(0, 0), A(4, 0) and B(0, 4). The values of Z at these points are as follows

Therefore, the maximum value of Z is 16 at the point B(0, 4).

Construct the following table of values of objective function

(i) (d): Minimum value of Z is -48 which occurs at (0,8).
(ii) (a): Maximumvalue of Z is 20, which occurs at (5,0).
(iii) (b): Maximum of Z - Minimum of Z
= 20 - (-48) = 20 + 48 = 68
(iv) (c): The corner points of the feasible region are O(0,0), A(3, 0), B(3, 2), C(2, 3), D(0, 3).
(v) (d)