Our problem is to minimise and maximise
Z = x + 2y                     ....(i)
Subject to constraints
\(\begin{aligned} & x+2 y \geq 100 \end{aligned}\)        ....(ii)
\(\begin{aligned} 2 x-y \leq 0 \end{aligned}\)            ....(iii)
\(\begin{aligned} 2 x+y \leq 200 \end{aligned}\)        ....(iv)
and \(\begin{aligned} x \geq 0, y \geq 0 \end{aligned}\)     ....(v)
Table for line x + 2y = 100 is

So, the line x + 2y = 100 is passing through the points (0, 50) and (100, 0).
On putting (0, 0) in the inequality x + 2y \(\geq\) 100, we get
0 + 2 \(\times\) 0 \(\geq\) 100
\(\Rightarrow\)     0 \(\geq\) 100 (which is false)
So, the half plane is away from the origin.
Table for line 2x - y = 0 is

So, the line 2x - y = 0 is passing through the points (0, 0) and (10, 20).
On putting (5, 0) in the inequality 2x - y \(\leq\) 0,
we get
2 \(\times\) 5 - 0 \(\leq\) 0
\(\Rightarrow\)     10 \(\leq\) 0 (which is false)
So,the half-plane is towards Y-axis.
Table for line 2x + y = 200 is

So, the line 2x + y = 200 is passing through the points (0, 200) and (100, 0).
On putting (0, 0) in the inequality 2x + y \(\leq\) 200, we get 2 \(\times\) 0 + 0 \(\leq\) 200 \(\Rightarrow\) 0 \(\leq\) 200 (which is true).
So, the half plane is towards the origin.
Also, x, y \(\geq\) 0.
So, the region lies in the lst quadrant.

Clearly, feasible region is ABCDA.
On solving equations 2x - y=0 and x + 2y=100,
we get B(20, 40).
Again, solving the equations 2x - y = 0 and 2x + y = 200, we get C(50, 100).
The corner points of the feasible region are A(0, 50), B(20, 40), C(50, 100) and D(0, 200).
The values of Z at corner points are given below

The maximum value of Z is 400 at D(0, 200) and the minimum value of Z is 100 at all the points on the line segment joining A(0, 50) and B(20, 40).

Maximise Z = -3x - 5y
Subject to the constraints
\(\begin{aligned} -2 x+y \leq 4, x+y \geq 3, \end{aligned}\)
\(x-2 y \leq 2 \text { and } x \geq 0, y \geq 0 \)
Now, considering the inequations as equations, We get
            -2x + y = 4           ....(i)
               x + y = 3           ....(ii)
and        x - 2y = 2           ....(iii)
Table for line -2x + y = 4 is

So, it passes through the points (-2, 0) and (0, 4) On putting (0, 0) in the inequality -2x + y \(\leq\) 4, we have
0 \(\leq\) 4 (which is true)
So, the half plane is towards the origin.
Table for line x + y = 3 is

So, it passes through the points (3, 0) and (0, 3).
0 + 0 \(\geq\) 3 (which is false)
So, the half plane is away from the origin.
Table for line x - 2y = 2 is

So, it passes through the point (2, 0) and (0, -1).
On putting (0, 0) in the inequality x - 2 y \(\leq\) 2.
0 \(\leq\) 2 (which is true)
So, the half plane is towards the origin.
Also, x \(\geq\) 0, y \(\geq\) 0, so the feasible region lies in the Ist quadrant.
The point of intersection of Egs. (i) and (ii) is
\(\left(-\frac{1}{3}, \frac{10}{3}\right)\), Eqs. (ii) and (iii) is \(\left(\frac{8}{3}, \frac{1}{3}\right)\) and Eqs. (i) and (iii) is \(\left(\frac{-10}{3}, \frac{-8}{3}\right)\). The graphical representation of the above system of inequations is given below.

\(\therefore\) Clearly, feasible region is shaded.
The value of Z at corner points are as follows

Here, the feasible region is unbounded and the open half plane determined by -3x - 5y > \(\frac{-29}{3}\) has no point in common with the feasible region. Hence, \(Z=-\frac{29}{3}\) is maximum at \(A\left(\frac{8}{3}, \frac{1}{3}\right)\)

Given, Z = x + 2y
Subject to the constraints
\(2 x+y \geq 3 ; x+2 y \geq 6 \text { and } x \geq 0, y \geq 0\)
Now, considering the inequations as equations, we get
2x + y = 3             ....(i)
x + 2y = 6             ....(ii)
Table for line 2x + y = 3 is

On putting (0, 0) in the inequality \(2 x+y \geq 3\) 0 \(\geq 3\) (which is false)
So, the half plane, is away from the origin
Table for line x + 2y = 6

On putting (0, 0) in the inequality x + 2y \(\geq\) 6 0  \(\geq\) 6 (which is false)
So, the half plane is away from the origin.
Also, x \(\geq\) 0 and y \(\geq\) 0, so the feasible region lies in the Ist quadrant.
The point of intersection of Eqs. (i) and (ii) is (0, 3).
The graphical representation of the above system of inequations is given below.

\(\therefore\) Feasible region is shaded region above.

Here, the feasible region is unbounded and the open half plane determined by x + 2y < 6 has no point in common with the feasible region. Hence, Z = 6 is minimum at A(0, 3) and C(6, 0) and also on the line segment AC.

We have, maximise,Z = 300 x + 600 y            ...(i)
Subject to the constraints, \(x+2 y \leq 12\)     ...(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 + 2y \(\leq\) 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 2x + 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 4x + 5y \(\geq\)20, we get
4(0) + 5(0) \(\geq\) 20 \(\Rightarrow\) 0 \(\geq\) 20
which is not true.
So, the half plane is away from the origin
Also, x. y \(\geq\)0
So, the region lies in first 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 values of Z at corner points are given below.

The maxinum value of Z is 3600 at D(4, 4) and E(0, 6).

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 minimise
Z = 5x + 10 y                             ...(i)
Subject to constraints
\(\begin{aligned} x+2 y & \leq 120 \end{aligned}\)                  ...(ii)
\(\begin{aligned} x+y & \geq 60 \end{aligned}\)                     ...(iii)
\(\begin{aligned} x-2 y & \geq 0 \end{aligned}\)                     ...(iv)
and \(x \geq 0, y \geq 0\)
Table for line x + 2y = 120 is

On putting (0, 0) in the inequality x + 2y \( \leq\) 120, we get
0 + 2 \(\times\) 0 \( \leq\) 120
\(\Rightarrow\)         0  \( \leq\) 120  (which is true)
So, the half plane is towards the origin. Secondly, table for the line x + y = 60 is

On putting (0, 0) in the inequality x + y \(\geq\) 60, we get
0 + 0 \(\geq\) 60
\(\Rightarrow\)     0 \(\geq\) 60 (which is false)
So, the half plane is away from the origin. Table for the line x - 2y = 0 is

On putting (5, 0) in the inequality x - 2y \(\geq\) 0, we get
5 - 2  \(\times\) 0 \(\geq\) 0
\(\Rightarrow\)     5 \(\geq\) 0     (which is true)
Thus, the half plane is towards the X-axis.
Since, x, y \(\geq\) 0
\(\therefore\) The feasible region lies in the Ist quadrant.

Clearly, feasible region is ABCDA.
On solving equations x - 2 y = 0 and x + y = 60, we get D(40, 20) and on solving equations x - 2y = 0 and x + 2y = 120, we get C (60, 30). The corner points of the feasible region are A (60, 0), B (120, 0), C (60, 30) and D (40, 20). The values of Z at these points are as follows

Clearly, the minimum value of Z is 300 at the point A(60, 0).

Given, Z = -3x + 4y
Subject to the constraints
x + 2y \( \leq\) 8; 3x + 2y \( \leq\) 12 and x \(\geq\)0, y \(\geq\) 0
Now, considering the inequations as equations, we get
  x + 2y = 8                  ...(i)
3x + 2y = 12                ...(ii)
Table for line x + 2y = 8 is

On putting (0, 0) in the inequality x + 2y \( \leq\) 8
0 \( \leq\) 8  (which is true)
So, half plane is towards the origin.
Table for line 3x + 2y = 12

On putting (0, 0) in the inequality 3x + 2y \( \leq\)12
0 \( \leq\) 12 (which is true)
So, half plane is towards the origin.
Also, x \(\geq\) 0 and y \(\geq\) 0, so the feasible region lies in the Ist quadrant.
The point of intersection of Eqs. (i) and (ii) is (2, 3).
The graphical representation of the above system of inequations is given below.

Hence, Z = -12 is minimum at (4, 0).

Let the number of large vans = x
and number of small vans = y
Minimise cost z = 400x + 200y
subject to constraints
\(200 x+80 y \geq 1200 \text { or } 5 x+2 y \geq 30\) ,
\(400 x+200 y \leq 3000 \text { or } 2 x+y \leq 15\)
\(x 

Suppose tailor A work for x days and tailor B work for y days.
The given data can be written in the tabular form as follows
\(\begin{array}{c|c|c|c} \hline \text { Tailor } & \begin{array}{c} \text { Number } \\ \text { of shirts } \end{array} & \begin{array}{c} \text { Number of } \\ \text { trousers } \end{array} & \text { Cost/day } \\ \hline A & 6 & 4 & Rs. 300 \\ B & 10 & 4 & \text { Rs. } 400 \\ \hline \text { Minimum requirement } & 60 & 32 & \\ \hline \end{array}\) 
Required linear programming problem is
Min (Z) = 300x + 400y
subject to constraints
\(6 x+10 y \geq 60\) 
\(4 x+4 y \geq 32\)  and \(x \geq 0, y \geq 0\) 

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).